What is the mass Cu produced if 9g of Cu2S reacts with 16 g O2? Help!

Cu2S+ O2-----> 2Cu + SO2

Question 2:
What is the percent yield if 24 g of I2 is formed when 130.0 g of HNO3 reacts with 285g of HI?
2HNO3 + 6HI-----> 3NO + 3I2 +4H2O

I am having my final exam on Tuesday for chemistry, and I trying to go back and refresh my memory. And I realized that I don't how to do those kind of problems!
Please Help
Thanks

1. Molecular Mass of Cu2S = 159.17g/mol
Molecular Mass of O2 = 32g/mol
Atomic Mass of Cu = 63.55g/mol

9g / 159.17g/mol = 0.056543318465 moles of Cu2S

16g / 32g/mol = 0.5moles of O2

Since the mole ratio of Cu2S to O2 is 1:1 we can say that Cu2S is our limiting reactant because it runs out first in our reaction.

Since the mole ratio of Cu2S to Cu is 1:2 our Cu moles will be double our Cu2S moles.

0.056543318465mols * 2 = 0.113086636929moles of Cu

0.113086636929mols * 63.55g/mol = 7.187grams of Cu

The answer is 7.187grams

2. First lets calculate what our ideal yield is (what we would get if this reaction went to absolute completion.)

Molecular mass of HNO3 = 63g/mol
Molecular mass of HI = 127.9g/mol
Molecular mass of I2 = 253.8g/mol

130g / 63g/mol = 2.06349206349moles of HNO3

285g / 127.9g/mol = 2.23354231975moles of HI

The mole ratio between HNO3 and HI simplifies to 1:3. From this we can say HI is our limiting reagent because it runs out first in the reaction.

The mole ratio between HI and I2 simplifies to 2:1. Therefore our I2 moles will be HALF of our HI moles.

2.23354231975moles / 2 = 1.11677115987moles of I2

1.11677115987mols * 253.8g/mol = 283.436520376grams of I2

Now to calculate percent yield we divide our experimental yield by our theoretical yield and multiply by 100.

(24g / 283.436520376g) * 100 =8.4675%

The answer if 8.4675%