Question

If 42.5 g N2 react with 10.1 g H2 according to the reaction below, how many grams of ammonia (NH3) can be produced, and how many grams of the excess reactant will be left over?

Unbalanced equation: N2 + H2 → NH3

Answer 1

this is what i have, i just want to make sure it's right

N2 + 3H2 -> 2NH3

42.5g N2 x [1 mol N2]/[28g N2] x [2 mol NH3]/[1 mol N2] x [17g NH3]/[1 mol NH3] = 51.6g NH3

10.1g H2 x [1 mol H2]/[2g H2] x [2 mol NH3]/[3 mol H2] x [17g NH3] x [1 mol NH3] = 57.2g NH3

42.5g N2 x [1 mol N2]/28g N2] x [3 mol H2]/[1 mol N2] x [2g H2]/[1 mol H2] = 9.1g H2

10.1g - 9.1g = 1g H2

51.6g of ammonia can be produced. There will be 1g of hydrogen gas left over.