For $N X_{2} + 3 H X_{2} 2 NH X_{3}$ , $14 g$ of $N X_{2}$ and $6 g$ of $H X_{2}$ are mixed and allowed to react completely. Taking $N = 14$ , $H = 1$ , what mass of $NH X_{3}$ is formed?

Answer: A

Moles:

N X_{2}

= 14/28 = 0.5,

H X_{2}

= 6/2 = 3.

H X_{2}

needed for 0.5 mol

N X_{2}

is 3 x 0.5 = 1.5 mol, less than available, so

N X_{2}

is limiting.

NH X_{3}

= 2 x 0.5 = 1 mol = 17 g (molar mass 17).