Name:_____________
Chemistry 112
Second Hour Exam
Remember- Show all work for partial credit
1. (10 points) Balance the following chemical equations:
A. Fe(s) + O2(g) 6 Fe2O3(s)
4Fe(s) + 3O2 (g) 62Fe2O3(s)
B. Eu(s) + HF (g) 6 EuF3(s) + H2(g)
2 Eu(s) + 6 HF(aq) 6 2EuF3(s) + 3H2(g)
2. (10 points) Silicon is produced by the following chemical reaction:
SiO2 (s) + C(s) 6 Si(s) + CO(g)
If I start with 20 g of SiO2 and an excess of C, how many grams of Si
will be produced?
Molar mass of SiO2 = 28.08 +2(16.00) = 60.08g/mole
20 g SiO2 ×
1 mole SiO2
1 mole Si
28.08 g Si
×
×
= 9.35g Si
60.08 g SiO2 1 mole SiO2 1 mole Si
2
3. (15 points) Sodium metal reacts violently with water in the reaction:
2Na(s) + 2H2O(l) 6 2Na+ (aq)+ H2(g) +2OH-(aq)
A. If I have 2 grams of Na metal and 2 grams of water, which is the
limiting reagent.
If 2 g Na:
2 g Na × 1 mole Na/22.99g Na ×1 mole H2/2 mole Na = .0435 mole H2
If 2 g H2O
2 g H2O × 1 mole H2O /18.0g Na ×1 mole H2/2 mole H2O
= .0556 mole H2
Na gives the smaller amount of product so it must be the limiting reagent
B. If my yield of H2(g) is .07 g, what is the % yield of the above reaction
% = actual/theoretical x 100%
= .07/(.0435 mole H2 × 2.02g g/mol)× 100% = 79.7%
4. (9 points) Give an example of
(A) a strong electrolyte HCl, NaOH, NaCl, etc.
(B) a weak electrolyte Acetic Acid, ammonia, etc
(C) a nonelectrolyte Sucrose, ethanol, etc
5. (10 points) In lab we used 10 mL of 6M Sulfuric acid to change 0.25 g of
NaC2O4 to H2C2O4. How many moles of sulfuric acid was this?
10 ml × (1 l/1000 ml) × (6 moles/l) = .06 moles
3
6. (9 points) For a lecture demonstration I showed that K2CrO4(aq) reacts
with Ba(NO3)2(aq) to form a precipitate.
A. Write the molecular equation for this reaction
K2CrO4(aq) + Ba(NO3)2(aq) 62KNO3(aq) + BaCrO4(s)
B. Write the complete ionic equation for this reaction
2K+(aq) + CrO42-(aq) + Ba2+(aq) + 2NO3-(aq) 6
2K+(aq) +2NO3-(aq) + BaCrO4(s)
C. Write the net ionic equation for this reaction
Ba2+(aq) + CrO42-(aq) 6 BaCrO4(s)
7.(9 points) Are the following compounds soluble (S) insoluble (I) or slightly
soluble (SS)
KNO3____S_______ Ca3(PO4)2___SS_____ Fe(NO3)3____S_____
NaOH____S________ BeSO4_____S______ BaOH____SS______
PbCl 2_____I_______ (NH4)2SO4____S____ Li 2S_____S_______
8.(10 points) How much 0.103 M NaOH is required to neutralize 32.6 mL of
0.0957 M HCl
In Neutralization moles of acid = mole of base
Moles of acid = molarity × volume = 32.6 ml × (1L/1000ml) × .0957 mol/l
= 3.12x10-3 moles
Moles base = 3.12x10-3 mole = molarity × volume
3.12x10-3 moles = .103 Mole/L × X volume
X = 3.12x10-3 moles/(.103 mole/l)
3.029x10-2 l
=30.29 mL
4
9. Given the following unbalanced redox reaction:
Mn2+ (aq) + NaBiO3(s) 6 Bi3+(aq) + MnO 4A. (8 points) Balance the reaction under acid conditions
Mn 6MnO44H2O + Mn2+ 6MnO44H2O + Mn2+ 6MnO4- + 8H+
4H2O + Mn2+ 6MnO4- + 8H+ + 5e2+
X2
NaBiO 36Bi3+
NaBiO 36Bi3++ Na+
NaBiO 36Bi3++ Na++ 3H2O
6H+(aq) + NaBiO 36Bi3++ Na++ 3H2O
2e- + 6H+(aq) + NaBiO 36 Bi3++ Na++ 3H2O
X5
8H2O + 2Mn2+ + 10e- + 30H+(aq) + 5NaBiO 3 62MnO 4-+ 16 H+ + 10e - + 5 Bi3++ 5 Na++ 15 H2O
2Mn2++ 5NaBiO 3 + 14 H+ 6 2MnO 4-+ 5 Bi3++ 5 Na++ 7 H2O
2Mn2+(aq) + 5NaBiO3 (s) + 14 H+(aq) 6 2MnO4-(aq)+ 5 Bi3+(aq)+ 5 Na+(aq)+ 7 H2O(l)
B. (4 points) How would I change the above reaction if the reaction
occurred under basic conditions
2Mn2++ 5NaBiO 3 + 14 H+ 6 2MnO 4-+ 5 Bi3++ 5 Na++ 7 H2O
+ 14OH+ 14 OH2Mn2++ 5NaBiO 3 + 14 H2O 6 2MnO 4-+ 5 Bi3++ 5 Na++ 7 H2O + 14 OH2Mn2+(aq)+ 5NaBiO3 (s)+ 7 H2O(l) 6 2MnO4-(aq)+ 5 Bi3+(aq)+ 5 Na+(aq)+ 14 OH-(aq)
C. (4 points) What is the reduction ½ reaction (Acid conditions)
2e- + 6H+(aq) + NaBiO 36 Bi3++ Na++ 3H2O
D. (4 points) Name the reducing agent in the reaction.
Mn2+