Please help with Chemistry !?

Question

I'm sooo stuck. I'd really appreciate a step by step if anyone would be so kind. Thanks in advance.

1. What is the maximum mass of lead that can be obtained by the reaction of 57.33 gramsof PbO and 33.80 grams of PbS?

2. What volume of 1.131M BaCl2 is required to react completely with 42.0 ML of .453 M NA2SO4 ?

3. If 3.00 mol of FeCl3 are ignited in the presence of 2.00 mol of O2 gas how much of which reagent is present in excess and therefore remains unreacted?

If anyone can help me i'd really appreciate it. I don't know what I am doing and this is worth 50 points which i reaaallyyy need!

Answer 1

1. PbO + PbS
moles PbO = 0.257 mol
mass of Pb from PbO: 0.257 mol * 207.2 g/mol = 53.25g
moles PbS = 0.141 mol
mass of Pb from PbS: 0.141 mol * 207.2 g/mol = 29.21g
total mass of Pb: 53.25g + 29.21g = 82.46g

2. BaCl2 + Na2SO4 ---> 2NaCl + BaSO4
moles Na2SO4: 0.042 L * 0.453 mol/L = 0.019 mol
- means you also need 0.019 moles BaCl2
volume BaCl2 required:
0.019 mol/ 1.131 mol/L = 0.0168 L or 16.8 mL

3. 4FeCl3 + 3O2 ---> 2Fe2O3 + 6Cl2
based from the balanced equation above, 4 moles of FeCl3 are needed for every 3 moles of O2 for the oxidation reaction. Hence, the excess reagent is FeCl3. Only 2.66 moles FeCl3 is needed to react 2 moles of O2. There is 0.33 mole of FeCl3 remains unreacted.

Answer 2

In any question like these, you first have to figure out a balanced chemical equation for the reaction.

Your first question confuses me, because I can't think of an equation or a reaction that would occur between these two compounds. Now, if the question is really "How much lead could you produce from 57.33 grams of PbO?", then I can deal with that. You first need to convert mass into moles. If you take the mass of PbO and divide by the molar mass of PbO, you then have moles of PbO. If you recognize that 1 mole of PbO contains one mol of Pb, then you can convert that many moles of PbO into grams of Pb by multiplying by the molar mass of Pb.

For your second question, you need to determine the balanced equation for this reaction. When you combine these solutions, BaSO4(s) is going to precipitate. In the balanced equation, you will have a ratio of 1 mol BaCl2 to 1 mol Na2SO4 to 1 mol BaSO4. So, using the volume (in liters) of the Na2SO4 solution and its molar concentration, you can determine how many moles of Na2SO4 you are starting with. You'll need exactly the same number of moles of BaCl2 to react completely with it. Finally by dividing that number of moles by the concentration of the BaCl2 solution, you'll have the volume (in Liters) of the barium solution.

Problem 3 is a standard Limiting Reactant question. Again, you need a balanced equation for the reaction. Once you have that, you can determine how many moles of the product you can form if you started with each of the reactants individually. The reactant which gives you the smallest number of moles of product is the limiting reactant, and the other is present in excess...

Hope this all helps...