Chemistry Textbook
- 1. Matter and MeasurementsToggle Dropdown
- 2. Atoms, Molecules and IonsToggle Dropdown
- 3. Composition of Substances and SolutionsToggle Dropdown
- 4. Stoichiometry of Chemical Reactions
- 5. ThermochemistryToggle Dropdown
- 6. GasesToggle Dropdown
- 7. Chemical Bonding and Molecular GeometryToggle Dropdown
- 8. Liquids, Solids, and Modern MaterialsToggle Dropdown
- 9. Solutions and Colligative PropertiesToggle Dropdown
- 10. KineticsToggle Dropdown
- 11. Chemical Equilibria and ApplicationsToggle Dropdown
- 12. ThermodynamicsToggle Dropdown
- 13. ElectrochemistryToggle Dropdown
- 14. AppendicesToggle Dropdown
- Appendix A: Periodic Table of Elements
- Appendix B: Essential Mathematics
- Appendix C: Units and Conversion Factors
- Appendix D: Fundamental Physical Constants
- Appendix E: Water Properties
- Appendix F: Composition of Commercial Acids and Bases
- Appendix G: Standard Thermodynamic Properties for Selected Substances
- Appendix H: Ionization Constants of Weak Acids
- Appendix I: Ionization Constants of Weak Bases
- Appendix J: Solubility Products
- Appendix K: Formation Constants for Complex Ions
- Appendix L: Standard Electrode (Half-Cell) Potentials
Chapter-4 Self Assessments
What is the coefficient of NO2 when the following reaction is balanced?
N4H6 + O2 → NO2 + H2O
Choose 1 answer
An equation is balanced when the same number of each element are represented on the reactant and product sides. View topic here.
Be sure to balannce the entire equation and then report the coefficient of Nitrogen. An equation is balanced when there same number of atoms of every element on the reactant and product side.. View topic here.
An equation is balanced when the same number of each element are represented on the reactant and product sides. View topic here
Nitrogen and hydrogen are balanced first. Balance oxygens by changing coefficient of O2. Double all coefficients to eliminate fractional coefficients.
Review topic here.
Review topic here.
An equation is balanced when the same number of each element are represented on the reactant and product sides. View topic here.
What are the coefficients when the following reaction is balanced?
H2SO4 + Fe(OH)3 → H2O + Fe2(SO4)3
Choose 1 answer
Check coefficient of H2 Review Topic
If a coefficient is not written, it means the coefficient is 1. Review Topic
Balance Fe. Then balance SO4 6 H+ ions from H2SO4 and 6 OH- ions from the Fe(OH)3 produce 6H2O.
Review Topic
Review Topic
The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen to produce hydrogen fluoride. What is the balanced equation for this reaction?
Choose 1 answer
Formulas for hydrogen and fluorine gases are wrong in the equation.
Formula for hydrogen gas is wrong in the equation
Fluorine and hydrogen gases are both diatomic gases, so, their formulas are H2 and F2. Hydrogen fluoride’s formula is HF
Formula for fluorine gas is wrong in the equation.
Formula for fluorine gas and hydrogen fluoride are wrong in the equation.
What are the coefficients of H3PO4 and CaCl2 when the following reaction is completed and balanced?
H3PO4(aq) + CaCl2(aq) →
Choose 1 answer
Double displacement reaction: 2H3PO4(aq) + 3CaCl2(aq) → 6HCl(aq) + Ca3(PO4)2(s)
H3PO4 has H+ and PO43- CaCl2 has Ca2+ and Cl- ions. In the double displacement reaction, ions switch places. H+ comes together with Cl- to form HCl; Ca2+ comes together with PO43- to form Ca3(PO4)2. Next, the equation is balanced.
H3PO4 has H+ and PO43- CaCl2 has Ca2+ and Cl- ions. In the double displacement reaction, ions switch places. H+ comes together with Cl- to form HCl; Ca2+ comes together with PO43- to form Ca3(PO4)2. Next, the equation is balanced.
Aqueous solutions of two ionic compounds are reacted. This will be a double displacement reaction. Identify the correct formulas of the two products and balance.
Aqueous solutions of two ionic compounds are reacted. This will be a double displacement reaction. Identify the correct formulas of the two products and balance.
Aqueous solutions of two ionic compounds are reacted. This will be a double displacement reaction. Identify the correct formulas of the two products and balance.
Aqueous solutions of two ionic compounds are reacted. This will be a double displacement reaction. Identify the correct formulas of the two products and balance.
Which of the following reaction will be a precipitation reaction?
Choose 1 answer
Acetic acid (CH3COOH) is an acid and sodium hydroxide (NaOH) is a base. The reaction between them would be a neutralization reaction that produces water and a soluble salt.
CH3COOH(aq) + NaOH(aq) → H2O (l) + CH3COONa (aq). This is not a precipitation reaction.
CH3COOH(aq) + NaOH(aq) → H2O (l) + CH3COONa (aq). This is not a precipitation reaction.
Cu(s) + HCl(aq) → CuCl2(aq) + H2(g). No precipitate is formed in the reaction. Also, Cu(s) + HCl(aq) reaction is a single displacement reaction! Precipitation reactions must be double displacement reactions.
Use solubility chart to determine if a precipitate will be formed or not.
The products are calcium sulfide and ammonium nitrate. From solubility rules, calcium sulfide is not soluble in water and separates out as a solid. This is a precipitation reaction.
Rusting of iron is an oxidation-reduction reaction. Rust is not a precipitate. A precipitation reaction is the reaction between two aqueous solution leading to the formation of a solid.
Hydrogen (H2) is produced according to the following equation:
2Cr + 2H3PO4 → 3H2 + 2CrPO4.
______ moles of hydrogen gas are produced when 0.4235 moles of chromium (Cr) react?
Choose 1 answer
The balanced equation tells the relationship between moles of Cr reacted to moles of H2 It suggests: “if” 2 moles Cr react, 3 moles H2 form.
Review Topic
Review Topic
The balanced equation tells the relationship between moles of Cr reacted to moles of H2 It suggests: “if” 2 moles Cr react, 3 moles H2 form.
Review Topic
Review Topic
4235 moles of Cr → moles of H2 using dimensional analysis.
Molar ratio from the balanced equation is the conversion factor:
3 mol Cr (reacts) = 2 mol H2 (forms)
Molar ratio from the balanced equation is the conversion factor:
3 mol Cr (reacts) = 2 mol H2 (forms)
The balanced equation tells the relationship between moles of Cr reacted to moles of H2 It suggests: “if” 2 moles Cr react, 3 moles H2 form.
Review Topic
Review Topic
The balanced equation tells the relationship between moles of Cr reacted to moles of H2 It suggests: “if” 2 moles Cr react, 3 moles H2 form.
Review Topic
Review Topic
In Haber-Bosch’s process, ammonia is synthesized according to the following reaction:
N2(g) + 3H2(g) → 2NH3(g)
How many nitrogen molecules are needed to react with 12 moles of hydrogen gas?
Choose 1 answer
Moles must be converted to molecules.
Review similar worked out textbook example-4.8.
Review similar worked out textbook example-4.8.
Use stoichiometric ration correctly. Also, convert moles to molecules.
Review similar worked out textbook example-4.8.
Review similar worked out textbook example-4.8.
Note that the question asks for nitrogen molecules (N2); not nitrogen atoms (not N)!
Review similar worked out textbook example-4.8.
Review similar worked out textbook example-4.8.
Check if molar ratio is used correctly. Review similar worked out textbook example-4.8.
Strategy: 12 mol H2 → mol N2 → molecules of N2
Conversion factors: stoichiometric ratio, Avogadro’s number
Review similar worked out textbook example-4.8.
Conversion factors: stoichiometric ratio, Avogadro’s number
Review similar worked out textbook example-4.8.
Silver is often extracted from ores as K[Ag(CN)2] and then recovered by the following reaction
2K[Ag(CN)2] (aq) + Zn(s) → 2 Ag(s) + Zn(CN)2(aq) + 2KCN(aq)
What mass of Zn(CN)2 is produced by the reaction of 35.27 g of K[Ag(CN)2]?
Choose 1 answer
Strategy: grams of K[Ag(CN)2] → moles of K[Ag(CN)2] → moles of Zn(CN)2 → grams of Zn(CN)2
Conversion factors: molar mass of K[Ag(CN)2]= 199 g/mol; stoichiometric ratio (2:1); molar mass of Zn(CN)2.
Review similar worked out textbook example-4.8.
Conversion factors: molar mass of K[Ag(CN)2]= 199 g/mol; stoichiometric ratio (2:1); molar mass of Zn(CN)2.
Review similar worked out textbook example-4.8.
Possible calculation error. Check strategy and calculation.
Review similar worked out textbook example-4.8.
Review similar worked out textbook example-4.8.
Note that molar ratio in the balanced equation related moles of substances. Grams must be converted to moles before using molar ratio.
Review similar worked out textbook example-4.8.
Review similar worked out textbook example-4.8.
Possible error in using molar mass. Note that molar mass is always the mass of 1 moles of substance.
Review similar worked out textbook example-4.8.
Review similar worked out textbook example-4.8.
Consider the reaction between hydrogen and iodine,
H2(g) + I2(g) → 2 HI(g).
What is the percent yield of the reaction that converts 1.0 kg of I2 to 0.87 kg of hydrogen iodide?
Choose 1 answer
See worked-out example- 4.15
Theoretical yield: 87 kg HI → g HI → mol HI → mol I2 → g H2 → kg H2
Percent yield = (experimental yield/theoretical yield) × 100
Percent yield = (experimental yield/theoretical yield) × 100
See worked-out example- 4.15
See worked-out example- 4.15
Convert kilograms to grams to begin the problem. See worked-out example- 4.15
Automotive air bags inflate when a sample of sodium azide, NaN3, is very rapidly decomposed.
2NaN3(s) → 2Na(s) + 3N2(g)
What mass of sodium azide is required to produce 73.6 L of nitrogen gas that has a density of 1.25 g/L?
Choose 1 answer
For pure substances (s, l, g), density and volume can be used to find mass of the substance which can then be turned to moles.
Possible error in converting volume to mass.
For pure substances (s, l, g), density and volume can be used to find mass of the substance which can then be turned to moles.
Volume of N2 (L) × Density of N2 (g/L) = grams of N2
grams of N2 → moles of N2 → moles of NaN3 → grams of NaN3
grams of N2 → moles of N2 → moles of NaN3 → grams of NaN3
For pure substances (s, l, g), density and volume can be used to find mass of the substance which can then be turned to moles.
Lithium nitride is a component of advanced batteries according to the following equation.
Li + N2 → Li3N (not balanced)
When 1.50 mol of lithium and 1.50 mol of nitrogen combine according to the above reaction, how many moles of product is formed and how many moles of the excess reagent is left over?
Choose 1 answer
Balance the equation and determine the limiting reagent. Review Topic
Balanced equation: 6Li + N2 → 2Li3
Li is the limiting reagent. 1.50 mol of Li produced 1.50/6×2 moles of Li3N. 1.50 mol of Li uses 1.50/6×1 mol N2, so, 1.25 mol are left.
Li is the limiting reagent. 1.50 mol of Li produced 1.50/6×2 moles of Li3N. 1.50 mol of Li uses 1.50/6×1 mol N2, so, 1.25 mol are left.
Balance the equation and determine the limiting reagent. Review Topic
Balance the equation and determine the limiting reagent. Review Topic
Balance the equation and determine the limiting reagent. Review Topic
Ammonia is oxidized to nitric oxide in the presence of a catalyst according to the following reaction:
4NH3 + 5O2 → 4NO + 6H2O
When 4.5 g of NH3 and 6.5 g of O2 react, how many grams of NO is formed?
Choose 1 answer
Limiting reagent was not determined correctly. Review Example 4.14
Grams must be converted to moles before determining the limiting reagent. Review example 4.14.
4.5 g of NH3 is 0.264 mol; 6.5 g O2 is 0.203 mol.
O2 is the limiting reagent. 6.5 g O2 produces 4.9 g NO.
Review example 4.14
O2 is the limiting reagent. 6.5 g O2 produces 4.9 g NO.
Review example 4.14
80.0 g of Aluminum is placed in a 1.00 L solution of 0.590 M sulfuric acid. Hydrogen gas and aluminum sulfate are produced in this reaction. How many grams of excess reactant remains after the reaction is complete?
Choose 1 answer
Identify the formulas of reactants and products and write a balanced equation.
Start by determining moles of Al and moles of sulfuric acid provided.
Review molarity equation and Limiting reagent concept video.
Start by determining moles of Al and moles of sulfuric acid provided.
Review molarity equation and Limiting reagent concept video.
Balanced equation: 2Al + 3H2SO4 → 3H2 + Al2(SO4)2
0 g of Al is 2.965 mol; 1.00 L of 0.590 M H2SO4 is 0.590 mol.
The limiting reagent is H2SO4. 0.590 mol H2SO4 uses 0.393 mol Al, so, 2.572 mol or 69.4 g is left.
0 g of Al is 2.965 mol; 1.00 L of 0.590 M H2SO4 is 0.590 mol.
The limiting reagent is H2SO4. 0.590 mol H2SO4 uses 0.393 mol Al, so, 2.572 mol or 69.4 g is left.
Looks like you calculated excess reagent used (10.6 g). One more step is needed. Calculate how much is left by subtracting amount used from the initial amount.
Review Concept Video.
Review Concept Video.
Identify the formulas of reactants and products and write a balanced equation.
Start by determining moles of Al and moles of sulfuric acid provided.
Review molarity equation and Limiting reagent concept video .
Start by determining moles of Al and moles of sulfuric acid provided.
Review molarity equation and Limiting reagent concept video .
Identify the formulas of reactants and products and write a balanced equation.
Start by determining moles of Al and moles of sulfuric acid provided.
Review molarity equation and Limiting reagent concept video.
Start by determining moles of Al and moles of sulfuric acid provided.
Review molarity equation and Limiting reagent concept video.
The reaction of aluminum with chlorine gas is shown below
2Al(s) + 3Cl2 (g) → 2AlCl3 (s)
10.0 g Aluminum reacted with 30.0 g of chlorine (Cl2) to produce 35.2 g of AlCl3.
What is the percent yield of the reaction?
Options- Correct:C
Choose 1 answer
Determine the limiting reagent first and calculate the theoretical yield of the product from grams or moles of the Limiting Reagent. Then find the percent yield. Here is an example
Limiting reagent was not determined correctly. Theoretical yield of AlCl3 must be calculated from the limiting reagent, not the excess reagent.
Here is an example
Here is an example
Cl2 is the limiting reagent. 30.0 g of Cl2 should produce 37.6 g AlCl3 (theoretical yield).
Percent yield = (experimental yield / theoretical yield) × 100.
Determine the theoretical yield from the limiting reagent.
Percent yield = (experimental yield / theoretical yield) × 100.
Determine the theoretical yield from the limiting reagent.
Determine the limiting reagent first and calculate the theoretical yield of the product from grams or moles of the Limiting Reagent. Then find the percent yield. Here is an example
Determine the limiting reagent first and calculate the theoretical yield of the product from grams or moles of the Limiting Reagent. Then find the percent yield. Here is an example
CO is used to obtain iron from iron ore according to the following reaction:
C0 + Fe2O3 → Fe + CO2 (not balanced)
The reaction of 92.7 g CO with 154 g of Fe2O3 produces 78.2 g of Fe. What is the percent yield of iron?
Choose 1 answer
Review balancing the equation and example 4.15
Review balancing the equation and example 4.15
Review balancing the equation and example 4.15
Balanced equation: 3C0 + 2Fe2O3 → 4Fe + 3CO2.
Fe2O3 is the limiting reagent. 154 g of Fe2O3 should produce 107.7 g of Fe. Since, only 78.2 g is produced, percent yield is 72.6%.
Review balancing the equation and similar example 4.15
Fe2O3 is the limiting reagent. 154 g of Fe2O3 should produce 107.7 g of Fe. Since, only 78.2 g is produced, percent yield is 72.6%.
Review balancing the equation and similar example 4.15
Limiting reagent was not determined correctly. Theoretical yield of Fe must be calculated from the limiting reagent, not the excess reagent.
Review example 4.15
Review example 4.15
Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines according to the following equation.
2 LiOH(aq) + CO2(g) → Li2CO3 (aq) + H2O(l)
How many grams of CO2 is absorbed when 500 mL of 2.5 M LiOH reacts?
Choose 1 answer
You probably did not convert volume to appropriate units. Review the molarity equation. Review Concept
You probably did not consider the molar ratio between LiOH and CO2. Review Concept
Possible error in units for molarity equation. Also, molar ratio 1:2 is not used. Review Concept
You probably did not convert moles of CO2 to grams. Review Concept
Strategy: First, we can calculate moles of LiOH present in 500 mL of 2.5 M LiOH solution using the molarity equation. Moles = Molarity ×Liters . Now, we can use the molar ratio to convert moles of LiOH to moles of CO2.
5 L × 2.5 mol/L = 1.25 mol LiOH → mol CO2 → grams of CO2
Review Concept
5 L × 2.5 mol/L = 1.25 mol LiOH → mol CO2 → grams of CO2
Review Concept
A 25.00 mL portion of 2.45 M solution of Aluminum nitrate is reacted with excess 0.400 M sodium carbonate solution. What is the mass of the precipitate formed?
Hint: Write the balanced equation.
Choose 1 answer
Write the balanced equation, using the solubility chart to identify the precipitate. Use stoichiometric ratio between aluminum nitrate and the precipitate.
Review units used in the molarity equation.
Balanced equation: 2Al(NO3)2 (aq) + 3Na2CO3(aq) → Al2(CO3)3(s) + 6NaNO3.
025 L × 2.45 mol/L = 6.125×10-2 mol Al(NO)3 → mol of Al2(CO3)3 → grams of Al2(CO3)3.
Sodium carbonate is the excess reagent. It’s molarity is not needed to solve the problem.
025 L × 2.45 mol/L = 6.125×10-2 mol Al(NO)3 → mol of Al2(CO3)3 → grams of Al2(CO3)3.
Sodium carbonate is the excess reagent. It’s molarity is not needed to solve the problem.
Write the balanced equation, using the solubility chart to identify the precipitate. Use stoichiometric ratio between aluminum nitrate and the precipitate.
Note that sodium carbonate is the excess reagent. It’s molarity is not needed to solve the problem.
What volume of 0.01050 M HCl solution is required to reach the end point to titrate 25.00 mL of a 0.01000 M Ca(OH)2 solution?
Ca(OH)2(aq)+2HCl(aq) → CaCl2(aq)+2H2O(l)
Choose 1 answer
Review rearranging molarity equation. Molarity x Liters; Liters = moles/Molarity
Strategy: 0.025 L × 0.01000 mol/L = 2.5×10-4 mol Ca(OH)2 → mol of HCl → L of HCl → mL of HCl.
Note: moles = Molarity x Liters; Liters = moles/Molarity
Note: moles = Molarity x Liters; Liters = moles/Molarity
Hint: Start by calculating moles of Ca(OH)2, followed by using stoichiometric ratio.
Use stoichiometric ratio from the equation to convert moles of Ca(OH)2 to moles of HCl.
Note: Do not use M1V1 = M2V2! This is not a dilution problem.
Note: Do not use M1V1 = M2V2! This is not a dilution problem.
Hint: Start by calculating moles of Ca(OH)2, followed by using stoichiometric ratio.
A sample of solid calcium hydroxide, Ca(OH)2, is allowed to stand in water until a saturated solution is formed. A titration of 75.00 mL of this solution with 5.00 × 10−2M HCl requires 6 mL of the acid to reach the end point.
What is the molarity of the saturated solution of Ca(OH)2?
Choose 1 answer
Equation: Ca(OH)2(aq)+2HCl(aq) → CaCl2(aq)+2H2O(l)
00 × 10−2mol/L × 0.0366 L = 1.83×10−3 mol HCl → mol Ca(OH)2 → Molarity of Ca(OH)2
Review Concept
00 × 10−2mol/L × 0.0366 L = 1.83×10−3 mol HCl → mol Ca(OH)2 → Molarity of Ca(OH)2
Review Concept
Looks like you did not use the stoichiometric ratio. Note: Do not use M1V1 = M2V2! This is not a dilution problem.
Hint: Start by using the molarity equation to find moles of HCl. Review example 4.16
Hint: Start by using the molarity equation to find moles of HCl. Review example 4.16
Check whether the stoichiometric coefficients were used correctly. Review example 4.16
HCl data: 00 × 10−2 M , 36.6 mL; Ca(OH)2 data: 75.00 mL
Hint: Use the molarity equation to find moles of HCl. Review example 4.16
Hint: Use the molarity equation to find moles of HCl. Review example 4.16
HCl data: 00 × 10−2 M , 36.6 mL; Ca(OH)2 data: 75.00 mL
Hint: Use the molarity equation to find moles of HCl. Review example 4.16
Hint: Use the molarity equation to find moles of HCl. Review example 4.16
A 1.02 g portion of an unknown iron(II) salt was titrated with 0.0185 M potassium permanganate (KMnO4) solution in the presence of an acid. 19.95 mL of potassium permanganate was needed to reach the end point. What is the % of iron (by mass) in the unknown iron sample?
The iron(II) ions and permanganate ions react according to the following equation:
5Fe2+(aq) + MnO4-(aq) + 8 H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
Choose 1 answer
Hint: mol KMnO4 → mol MnO4- → mol Fe2+ → grams Fe.
Hint: mol KMnO4 → mol MnO4- → mol Fe2+ → grams Fe.
First, calculate moles of KMnO4: 0185 mol/L × 0.01995 L = 3.691×10-4 mol KMnO4
Convert mol KMnO4 → mol MnO4- → mol Fe2+ → grams Fe.
% mass of Fe = (grams Fe/ 1.02g) × 100
To convert mol of KMnO4 to moles of MnO4-, use the 1:1 molar ratio between KMnO4 and MnO4- (because, 1 KMnO4 contains 1 MnO4-)
Convert mol KMnO4 → mol MnO4- → mol Fe2+ → grams Fe.
% mass of Fe = (grams Fe/ 1.02g) × 100
To convert mol of KMnO4 to moles of MnO4-, use the 1:1 molar ratio between KMnO4 and MnO4- (because, 1 KMnO4 contains 1 MnO4-)
Hint: mol KMnO4 → mol MnO4- → mol Fe2+ → grams Fe.
Hint: mol KMnO4 → mol MnO4- → mol Fe2+ → grams Fe.
In which of the following substances does sulfur have the highest oxidation state?
Choose 1 answer
The oxidation state of Sulfur in S4 is zero. Review Rules for Oxidation states.
The oxidation state of Sulfur in H2S is -2. Review Rules for Oxidation states.
The oxidation state of Sulfur in SO32- is +4. Review Rules for Oxidation states.
The oxidation state of Sulfur in H2SO4 is +6. Review Rules for Oxidation states.
The oxidation state of Sulfur in S2O62- is +5. Review Rules for Oxidation states.
Which of the following statements is correct?
Choose 1 answer
Reduction is gain of electrons. Review concept.
Increase in the oxidation number is equivalent to loss of electrons. The substance that loses electrons is said to be oxidized. Review concept.
Reduction is gain of electrons. Review concept.
Oxidation is loss of electrons. Review concept.
An oxidizing agent is the substance that is reduced in a reaction. Review concept.
Which element is reduced in the following reaction?
I - + MnO4- + H+ → I2 + MnO2 + H2O
Choose 1 answer
An element whose oxidation number decreased in a reaction is the one that is reduced because decrease in oxidation state corresponds to gain of electrons.
Iodine’s oxidation state increases from -1 to 0 in the reaction
Iodine’s oxidation state increases from -1 to 0 in the reaction
An element whose oxidation number decreased in a reaction is the one that is reduced because decrease in oxidation state corresponds to gain of electrons.
Manganese’s oxidation state decreases from +7 to +4 in the reaction.
Manganese’s oxidation state decreases from +7 to +4 in the reaction.
An element whose oxidation number decreased in a reaction is the one that is reduced because decrease in oxidation state corresponds to gain of electrons.
Oxygen’s oxidation state did not change during the reaction!
Oxygen’s oxidation state did not change during the reaction!
An element whose oxidation number decreased in a reaction is the one that is reduced because decrease in oxidation state corresponds to gain of electrons.
Hydrogen’s oxidation state did not change during the reaction!
Hydrogen’s oxidation state did not change during the reaction!
An element whose oxidation number decreased in a reaction is the one that is reduced because decrease in oxidation state corresponds to gain of electrons.
Which element is the reducing agent in the following reaction?
Fe2+ + H+ + Cr2O72- → Fe3+ + Cr3+ + H2O
Choose 1 answer
Iron is oxidized (+2 to +3) and will be the reducing agent.
Hydrogen’s oxidation state did not change in the reaction.
An element that is oxidized in the reaction will be the reducing agent.
Chromium is reduced (+6 to +3) and is the oxidizing agent.
Chromium is reduced (+6 to +3) and is the oxidizing agent.
Oxygen’s oxidation state did not change in the reaction.
An element that is oxidized in the reaction will be the reducing agent. Hydrogen’s oxidation state did not change during the reaction.
Which element is the oxidizing agent in the following reaction?
Fe2S3 + 12HNO3 → 2Fe(NO3)3 + 3S + 6NO2 + 6H2O
Choose 1 answer
An element that is reduced in the reaction will be the oxidizing agent.
Iron has an oxidation state of +3 in both the reactant (Fe2S3) and the product (Fe(NO3)3). Since iron’s oxidation state did not change in the reaction, it is neither oxidized not reduced.
Iron has an oxidation state of +3 in both the reactant (Fe2S3) and the product (Fe(NO3)3). Since iron’s oxidation state did not change in the reaction, it is neither oxidized not reduced.
An element that is reduced in the reaction will be the oxidizing agent.
Sulfur has an oxidation state of -2 in the reactant (Fe2S3) and 0 in the product (S). Sulfur is oxidized in the reaction.
Sulfur has an oxidation state of -2 in the reactant (Fe2S3) and 0 in the product (S). Sulfur is oxidized in the reaction.
An element that is reduced in the reaction will be the oxidizing agent.
Hydrogen has an oxidation state of +! in both the reactant (HNO3) and the product (H2O). Since hydrogen’s oxidation state did not change in the reaction, it is neither oxidized not reduced.
Hydrogen has an oxidation state of +! in both the reactant (HNO3) and the product (H2O). Since hydrogen’s oxidation state did not change in the reaction, it is neither oxidized not reduced.
Nitrogen has an oxidation state of +5 in the reactant (HNO3) and +4 in the product (NO2). Nitrogen is reduced in the reaction and is the oxidizing agent.
An element that is reduced in the reaction will be the oxidizing agent.
Oxygen’s oxidation state did not change in the reaction!
Oxygen’s oxidation state did not change in the reaction!