14.0g n2 react with 3.15g h2 to give an actual yield of 14.5 nh3

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%

85.3%  

The answer is d.potassium