Limiting reactants and theoretical yield problems... help?

Question

I'm not so much interested in straight answers as much as I'm interested in the work that's involved in figuring out these problems. D: I need to know how to get from one point to another on problems like these if I have a hope of passing the test.

1. In the reaction shown below 4.0 mol of NO is reacted with 4.0 mol of O2

2NO + O2 ---> 2NO2

a. Which is the excess reactant, and which is the limiting reactant?
b. What is the theoretical yield, in units of mol, of NO2?

2. In the reaction shown below 64g CaC2 is reacted with 64g H2O.

CaC2(s) + 2H2O(l) ---> C2H2(g) + Ca(OH)2(s)

a. Which is the excess reactant, and which is the limiting reactant?
b. What is the theoretical yield of C2H2?
c. What is the theoretical yield of Ca(OH)2?

3. In the reaction shown below, 28g of nitrogen are reacted with 28g of hydrogen.

N2(g) + 3H2(g) ---> 2NH3(g)

a. Which is the excess reactant, and which is the limiting reactant?
b. What is the theoretical yield of ammonia?
c. How many grams of the excess reactant remain?

Almost forgot to say thanks in advance!=0

Answer 1

1. The coefficients of the reactants (2 mol of NO react with 1 mol of O2) tell you the required ratio. You have 4 mol of NO so you need 2 mol of O2...and you have 4, an excess.
a. excess is O2 so NO is limiting
b. again the coefficients tell you that you will get 2 mol NO2 for 2 mol NO. Since NO is limiting and you have 4 mol, the theorectical yield of NO2 is 4 mol.

2. This is like problem 1 but you were given grams, not mol. You need to convert to mol-
64g CaC2/ 64.1 g/mol = moles of CaC2 (1.00 mol)
64g H2O/18.0 g/mol = moles of H2O (3.56 mol)
1 mol of CaC2 reacts with 2 mol H2O
a. You need 2 mole H2O, and have 3.56 mole; H2O in excess
so CaC2 is limiting
b. You expect 1 mol C2H2 for each mol CaC2 (1:1)
c. You expect 1 mol of Ca(OH)2 for each mol of CaC2 (1:1)

3. The same type of question as #2 with an added wrinkle.
In #2 you needed 2.o mol of H2O and you had 3.56 mol. After the reaction you would have 3.56-2.00 = 1.56 mol of H2O left and 1.56 mol H2O x 18 g/mol = 28.1 g H2O left.
Now you do #3

REMEMBER: On your test you may have to first write the BALANCED equation. The coefficients of the balanced equation tell you the molar ratio of the reactants and products.

Answer 2

For the first type of problem, you need to have the balanced equation and the number of moles of each reactant. (The equation is already balanced in this case).

From the equation, you can see that it takes 2 NO's to react with 1 O2. Since you have 4 moles of each reactant, you can easily see that the NO will run out when only 1/2 the oxygen is gone. So NO is the limiting reactant and oxygen is the excess reactant.

To get the theoretical yield, just calculate how much product would be created at the point where the limiting reactant is used up.

In this case, the ratio of NO to NO2 is 1 to 1, so the moles of product formed would be equal to the moles of limiting reactant which is 4 moles

For #2:
You need to balance the equation:

CaC2 + 2 H2O ---> C2H2 + Ca(OH)2

Now, you need to calculate moles of reactants:

64g/[(40.08) + (2 x 12.01)]g/mole = 1.0 mole CaC2

64g/[16.00 + (2 x 1.01)]g/mol = 3.55 moles H2O

Now, use the ratios from the balanced reaction and the number of moles of each reactant to see which one will run out first:

You can see that you need 2 moles of water for each mole of CaC2 and the number of moles of CaC2 is 1.0, so you would only need 2 moles of water, to react with all the Calcium carbide.

The theoretical yield of C2H2 (in moles) happens to be the same number of moles as the limiting reactant, CaC2, as from the balanced equation, they are in a 1 to 1 ratio. So, one mole of CaC2 reacted, one mole of acetylene (C2H2) created. So 1.0 moles or:

[(2 x 12.01) + (2 x 1.01)]g/mole x 1.0mole; = 26g of C2H2 is created.

The theoretical yield of Ca(OH)2 is the same number of moles as the CaC2 that reacted, as they are in a 1 to 1 ratio from the balanced equation, so:

1.0 mole x [40.08 + (2 x 16.00) + (2 x 1.01)]g/mole = 74.1g

Look, I don't have time to do all your homework; for the last question, you can do it the same way, balance the equation, then determine number of moles of each of the reactants. You can detemine which reactant will run out first as we did before. Then you can find the moles of ammonia that could be created untill one of the reactants is gone (remember to use the ratio from the balanced equation). For the final part, calculate total moles of the excess reactant, then subtract the amount that got used up to get the answer.