Percent of moles and grams

Question

How would i go about solving the following question?

When 10.0 g of NH3reacts, the actual yield of N2is 8.50 g. What is the percent yield?

4NH3(g) + 6NO(g) →5N2(g) + 6H2O(l

Answer 1

We have to figure out the theoretical yield:

10 g NH3 * (1 mol NH3 / 17 g NH3) * (5 mol N2 / 4 mol NH3) * (28 g N2 / 1 mol N2) = 20.59 g N2 is the maximum amount we could get.

To find percent yield, take the actual yield and divide by theoretical yield:

8.5 g / 20.59 g = 0.413 --> 41.3%