You can determine the limiting reactant and the percentaje yield as follow.
1) State the equation that represents the chemical reaction
N2 (g) + 3H2 (g) → 2NH3 (g)
molar ratios: 1 mol N2 : 3 mol H2 : 2 mol NH3
2) Convert the two masses to moles by dividing each by its molar mass
14.0 g N2 / 28 g/mol = 0.5 mol
3.15 g H2 / 2g/mol = 1.575 mol
Proportion = 0.5 mol N2 / 1.575 mol H2 = 0.317 mol N2 / mol H2
3) Given that the theoretical ratio 1 / 3 > actual proportion 0.317 , the H2 is the limiting reactant (it will be consumed and yet some moles of N2 will remain without reacting).
4) Percentage yield = actual yield * 100 / theoretical yield
Theoretical yield = [ 2 mole NH3 * 1 mole N2 ] * 0.5 mol N2 = 1 mole NH3
actual yield = 14.5 g NH3 / molar mass NH3 = 14.5 g / 17 g/mol = 0.853 mol
Percentage yield = actual yield * 100 / theoretical yield = 0.853 mol * 100 / 1 mol = 85.3%
Answer: 85.3%
