Answers Homework from Tuesday

ANSWERS HOMEWORK FROM TUESDAY

Section 4.1 Chemical Equations p. 112

Practice Problems p. 112

1.  a) calcium + fluorine ® calcium fluoride

reactants product

b) barium chloride + hydrogen sulfate ® hydrogen chloride + barium sulfate

reactants products

c) calcium carbonate + carbon dioxide + water ® calcium hydrogencarbonate

reactants product

d) hydrogen peroxide ® water + oxygen

reactants products

e) sulfur dioxide + oxygen ® sulfur trioxide

reactants products

2. 

Practice Problems p. 113

3.  a) Zn(s) + Cl2(g) ® ZnCl2(s)

b) Ca(s) + 2 H2O(l) ® Ca(OH)2(s) + H2(g)

c) Ba(s) + 2 S(s) ® BaS2(s)

d) Pb(NO3)aq + Mg(s) ® Mg(NO3)2(aq) + Pb(s)

4.  a) CO2(g) + CaO(s) ® CaCO3(s)

b) 4 Al(s) + 3 O2(g) ® 2 Al2O3(s)

c) 2 Mg(s) + O2(g) ® 2 MgO(s)

Practice Problems p. 116

5.  a) S(s) + O2(g) ® SO2(g)

b) P4(s) + 5 O2(g) ® P4H10(s)

c) H2(g) + Cl2(g) ® 2 HCl(g)

d) SO2(g) + H2O(l) ® H2SO3(aq)

6. a) 4 Fe(s) + 3 O2(g) ® 2 Fe2O3(s) equation was balanced

b) 2 HgO(s) ® 2 Hg(l) + O2(g) equation was not balanced

c) 2 H2O2(aq) ® 2 H2O(l) + O2(g) equation was not balanced

Practice p. 117

7 a) 2 SO2(g) + O2(g) ® 2 SO3(g)

b) BaCl2(aq) + Na2SO4(aq) ® 2 NaCl(aq) + BaSO4(s)

8. P4(s) + 5 O2(g) ® P4O10(s)

P4O10(s) + 6 H2O(l) ® 4 H3PO4(aq)

Practice p. 118

9. a) As4S6(s) + O2(g)à As4O6(s) + SO2(g)

As4S6(s) + 9O2(g)à As4O6(s) + 6SO2(g)

b) Sc2O3(s) + H2O(l) à Sc(OH)3 (s)

Sc2O3(s) + 3H2O(l) à 2Sc(OH)3 (s)

c) C2H5OH(l) + O2(g) à CO2(g) + H2O (l)

2C2H5OH(l) + 6O2(g) à 4CO2(g) + 6H2O (l)

d) C4H10 (g) + O2(g)à CO(g) + H2O(g)

2C4H10 (g) + 9O2(g)à8CO(g) + 10H2O(g)

Section Review p. 118

1.  A chemical reaction is a reaction that occurs when one or more pure chemical entities react with each other to produce a new chemical entity.

2. a) sulfur dioxide + oxygen ® sulfur trioxide

SO2(g) + O2(g) ® SO3(g)

2 SO2(g) + O2(g) ® 2 SO3(g)

b) sodium metal + water ® hydrogen + sodium hydroxide

Na(s) + H2O(l) ® H2(g) + NaOH(aq)

2 Na(s) + 2 H2O(l) ® H2(g) + 2 NaOH(aq)

c) copper + nitric acid ® copper(II) nitrate + nitrogen dioxide + water

Cu(s) + HNO3(aq) ® Cu(NO3)2 + NO2(g) + H2O(l)

Cu(s) + 4 HNO3(aq) ® Cu(NO3)2 + 2 NO2(g) + 2 H2O(l)

3.  H2O2(aq) ® H2O(l) + O2(g) cannot be balanced by changing O2 to O, because then you would be changing the identity of a product.

4.  a) 4 Al(s) + 3 O2(g) ® 2 Al2O3(s)

b) 2 Na2S2O3(aq) + I2(aq) ® 2 NaI(aq) + Na2S4O6(aq)

c) 2 Al(s) + Fe2O3(s) ® Al2O3(s) + 2 Fe(s)

d) 4 NH3(g) + 5 O2(g) ® 4 NO(g) + 6 H2O(l)

e) Na2O(s) + (NH4)2SO4(aq) ® Na2SO4(aq) + H2O(l) + 2 NH3(aq)

f) C5H12(l) + 8 O2(g) ® 5 CO2(g) + 6 H2O(g)

5.  The reaction equation in this problem is:

Fe(s) + CuSO4(aq) ® Cu(s) + FeSO4(aq)

-  The total mass of reactants, water and container was: 0.58 + 1.600 + 100.00 + 40.32 = 142.5 g

-  The total mass of products, water and container is: 142.5 g

-  No mass is lost in a chemical reaction (law of conservation of mass). Had one of the products been a gas, then the gas could have escaped the container and that would make it look as if mass had been lost.