Question

Given 6.0 mol of N2 are mixed with 12.0 mol of H2. Equation: N2 + 3H2--> 2NH3

Which chemical is in excess? What is the excess in moles?
Theoretically, how many moles of NH3 will be produced?
If the percentage yield of NH3 is 80%, how many moles of NH3 are actually produced?

Answer 1

1) Excess reagent

1 mol N2 / 3 mol H2

6.0 mol N2 *3 mol H2 / 1 mol N2 = 18 mol H2

18mol H2 > 12 mol H2 => H2 is limiting (you need 18 mol H2 to use all the 6 mol N2), then N2 is in excees.

12.0 mol H2 * 1mol N2/ 3 mol H2 = 4 mol N2 is the quantity that will react, then the excess is 6 mol N2 - 4 mol N2 = 2 mol N2

2) NH3 produced

12 mol H2 * [2 mol NH3 / 3 mol H2] = 8 mol NH3

Aslso, 4 mol N2 *[2molNH3 / 1 molN2] = 8 mol NH3, the same result.

3) Yield

80% * 8 mol NH3 = 6.4 mol NH3