Question

1.29 g H2 is allowed to react with 9.77 g N2, producing 2.03 g NH3.

Part A

What is the theoretical yield in grams for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

Part B

What is the percent yield for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

Answer 1

N2 + 3H2 = 2NH3

change to mol as Mol = mass/MW

mol H2 = 1.29/2 = 0.95

mol of N2 = 9.77/28 = 0.3489

mol of NH3 = 2.03/17 = 0.119411

theoretical yield = moles that shoudl be formed with 100% reaction

1 mol of N2 reacts with 3 mol of h2 to form 2 mol of nh2

then

0.95 mol of h2 need 0.95/3 = 0.316666 mol of N2

we have 0.3489 mol of N2, so H2 is limiting

0.95 mol of H2 will form 2/3*0.95 = 0.633333 mol of NH3

theoreticla yield = 0.633333

real yield = 0.119411

% yield = real/theoretical * 100 = 0.119411/0.633333 *100 = 18.85%