We're given the following reaction:

**3 H _{2}(g) + N_{2}(g) → 2 NH_{3}(g)**

a) We're being asked to calculate the moles of NH_{3} that can be produced from 12.0 mol of H_{2} and excess N_{2}.

From the balanced reaction: 3 moles of H_{2} produces 2 moles NH_{3}.

moles H_{2} (mole-to-mole comparison) → moles NH_{3}

$\mathbf{moles}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{12}{\mathbf{}}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}\mathbf{}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}{\mathbf{3}{\mathbf{}}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}}$

Hydrogen gas, H_{2}, reacts with nitrogen gas, N_{2}, to form ammonia gas, NH_{3}, according to the equation

3 H_{2}(g) + N_{2}(g) → 2 NH_{3}(g)

Use molar masses expressed to five significant figures throughout.

a. How many moles of NH_{3} can be produced from 12.0 mol of H_{2} and excess N_{2}? Express your answer numerically in moles.

b. How many grams of NH_{3} can be produced from 4.10 mol of N_{2} and excess H_{2}? Express your answer numerically in grams.

c. How many grams of H_{2} are needed to produce of NH_{3}? Express your answer numerically in grams.

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