# 7.E: Chapter 7 Homework Answers

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Evidence of a Chemical Reaction

1. a) b)

3. c)

Writing and Balancing Chemical Equations

5. b)

7.

a) Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

b) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

c) Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l)

d) 2Fe2O3(s) + 6Cl2(g) → 4FeCl3(aq) + 3O2(g)

9.

a) N2O3(aq) + H2O(l) → 2HNO2(aq)

b) Xe(g) + 3F2(g) → XeF6(s)

c) NH4NO3(aq) → N2O(g) + 2H2O(l)

d) I2(s) + 6HNO3(aq) 2HIO3 + 6NO2(g) + 2H2(g)

11. C6H24(g) + 12O2(g) → 12H2O(l) + 6CO2(g)

13. C6H12O6 (aq) → 6C(s) + 6H2O(l)

15. 2H2O(g) + 2Na(s) → 2NaOH(s) + H2(g)

17.

a) 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)

b) 2MgS(s) + 3O2(g) → 2MgO(aq) + 2SO2(g)

c) 3P4S3(s) + 16KClO3(aq) → 6P2O5(aq) + 16KCl(aq) + 9SO2(g)

d) Fe3O4(aq) + 4CO(g) → 4CO2(g) + 3Fe(s)

19.

a) Correct.

b) Incorrect. You would need a 2 coefficient before HNO2(aq).

c) Incorrect. Rather than having a 2 coefficient on F2(g) and Cl2(g), you need to put a 2 coefficient on both NaCl(aq) and NaF(g) in order for this equation to be balanced.

d) Correct.

21.

a) H2O(l) → H2O(g)

b) H2O(l) → H2O(s)

c) H2O(g) → H2O (l)

Since all of the above reactions are only physical changes, they require no balancing because they are already balanced.

Solubility

23. No. The water molecules will still attract both the negative and the positive ions in the substance, but the polyatomic ions will break into units where the ion is intact. NO3- ions would not break down to Nitrogen and Oxygen ions.

Example:

Monoatomic Ion: KCl dissociates into K+ and Cl- ions

Polyatomic Ion: KNO3 dissociates into K+ and NO3- ions

25.

c) Na+

d) Mg2+

27.        BaSO4(aq)      CaCO3(aq)

29. a) strong electrolyte

31.

a) CaCrO4(aq) + Cu(NO3)2 (aq) → Ca(NO3)2 (aq) + CuCrO4(s)

b) 2KBr(aq) + Hg2SO4(aq) → K2SO4(aq) + Hg2(Br)2(s)

c) 2LiNO3(aq) + 2(NH4)2S(aq) → 2NH4NO3(aq) + Li2S(s)

d) Pb(NO3)2(aq) + 2NaCl(aq) → 2NaNO3(aq) + PbCl2(s)

33.

a) This reaction implies a phase change for both mixtures, which is incorrect.

Na2SO4(aq) + BaCl2(aq) → 2NaCl(aq) + BaSO4(s)

b) This reaction is not balanced. There should be a 2 coefficient before KI(aq) and KNO3(aq)

2KI(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbI2(s)

c) This reaction states that Ag2S is soluble when it is actually insoluble. It also states NaNO3 is insoluble when it is in fact soluble.

2AgNO3(aq) + Na2S(aq) → Ag2S(s) + 2NaNO3(aq)

35. Beaker A would form a precipitate because the reaction would be:

2K3PO4(aq) + 3Ca(NO3)2(aq) → 6KNO3(aq) + Ca3(PO4)2(s).

Beaker B would form a precipitate because the reaction would be:

2NaOH(aq) + BaBr2(aq) → Ba(OH)2(s) + 2NaBr(aq).

Ionic and Net Ionic Equations

37. A single displacement reaction involves the replacement of an element in a compound.

KCl(aq) + Na(s) → NaCl(aq) + K(s)

A double displacement reaction involves the replacement of an element in each compound. You ultimately switch the anions (or the cations) present in the reactants to get your products.

KCl(aq) + AgNO3(aq) → AgCl(s) + KNO3(aq)

39. a) Total Ionic: 2Rb+(aq) + 2F-(aq) + Cu2+(aq) + SO42-(aq)2Rb+(aq) + 2F-(aq) + Cu2+(aq) + SO42-(aq)

Net Ionic: No Reaction

b) Total Ionic: Sr2+(aq) + 2Br-(aq) + 2K+(aq) + SO42-(aq) → SrSO4(s) + 2K+(aq) + 2Br-(aq)

Net Ionic: Sr2+(aq) + SO42-(aq) → SrSO4(s)

c) Total Ionic: Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq)Na+(aq) + Cl-(aq) + H2O(l)

Net Ionic:  H+(aq) +OH-(aq) → H2O(l)

d) Total Ionic: 2Bi(OH)3(s) + 6H+(aq) + 3SO42-(aq) → 2Bi3+(aq) + 3SO42-(aq) + 6H2O(l)

Net Ionic: 2Bi(OH)3(s) + 6H+(aq) → 2Bi3+(aq) + 6H2O(l)

Acid-Base Reactions

41. Neutralization, water, salt

43. a) 2HClO3(aq) + Zn(OH)2(s) → Zn(ClO3)2(aq) + 2H2O(l)

b) H2SO4(aq) + Ba(OH)2(s) → BaSO4(s) + 2H2O(l)

c) HBr(aq) + LiOH(aq) → LiBr(aq) + H2O(l)

d) 2HF(aq) + Ca(OH)2(aq) → CaF2(s) + 2H2O(l)

Combustion Reactions

45. a) 2C6H6(l) + 15O2(g) → 6H2O(l) + 12CO2(g)

b) C6H12O6(aq) + 6O2(g) → 6H2O(l) + 6CO2(g)

c) C2H5OH(l) + 3O2(g) → 3H2O(l) + 2CO2(g)

Classifying Chemical Reactions

47. a) Double Displacement           b) Decomposition         c) Single Displacement       d) Synthesis

49. a) Decomposition                         b) Double Displacement                          c) Single Displacement

51. a) Decomposition                         b) Single Displacement                          c) Single Displacement

Cumulative Challenge Problems

53.

a) Molecular: BaCl2(aq) + Na2SO4(aq) → 2NaCl(aq) + BaSO4(s)

Total Ionic: Ba2+(aq) + 2 Cl-(aq) + 2Na+(aq) + SO42-(aq) → 2Na+(aq) + 2Cl-(aq) + BaSO4(s)

Net Ionic: Ba2+(aq) + SO42- (aq) → BaSO4(s)

b) Molecular:  Na2S(aq) + 2LiF(aq) → No Reaction

c) Molecular: H2SO4(aq) + Pb(NO3)2(aq) → PbSO4(s) + 2HNO3(aq)

Total Ionic: 2H+(aq) + SO42-(aq) + Pb2+(aq) + 2NO3-(aq) → PbSO4(s) + 2H+(aq) + 2NO3-(aq)

Net Ionic: Pb2+(aq) + SO42-(aq) → PbSO4(s)

55. a) Gas Evolution Reaction

Molecular: 2HClO4(aq) + CaCO3(s) → H2O(l) + CO2(g) + Ca(ClO4)2(aq)

Sum of Coefficients: 6

b) Precipitation Reaction

Molecular: 2Fe(s) + 3Cu(NO3)2(aq) → 3Cu(s) + 2Fe(NO3)3(aq)

Sum of Coefficients: 10

c) Combustion Reaction

Molecular: C9H20(l) + 14O2(g) → 10H2O(l) + 9CO2(g)

Sum of Coefficients: 34

57. a) 2(CH3)2CHOH(l) + 9O2(g) → 8H2O(l) + 6CO2(g)

b) C10H8(s) + 12O2(g) → 4H2O(l) + 10CO2(g)

c) 2CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3(l) + 37O2(g) → 26H2O(l) + 24CO2(g)

d) C12H22O11(aq) + 12O2(g) → 11H2O(l) + 12CO2(g)

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