The balanced reaction for this will appear as:

N_{2} (g) + 3 H_{2} (g) → 2 NH_{3 }(g)

a. moles of NH_{3}:

$\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{16}\mathbf{.}\mathbf{5}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}\mathbf{}\left(\frac{2\mathrm{mol}{\mathrm{NH}}_{3}}{3\overline{)\mathrm{mol}{H}_{2}}}\right)$

mol NH_{3 }= 11 mol NH_{3}

a. How many moles of NH_{3} can be produced from 16.5 mol of H _{2} and excess N_{2}?

b. How many grams of NH_{3} can be produced from 3.78 mol of N _{2} and excess H_{2}.

c. How many grams of H _{2} are needed to produce 11.05 g of NH_{3} ?

d. How many molecules (not moles) of NH_{3} are produced from 2.92 × 10^{−4} g of H_{2}?

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