**Balanced Reaction: **

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

a. Moles of NH_{3} from 13.5 mol of H_{2 }and excess N_{2 }

Since N_{2} is excess, calculate directly from H_{2}.

$\mathbf{13}\mathbf{.}\mathbf{5}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}\mathbf{}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{2}}}{\mathbf{3}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}\mathbf{}}\mathbf{}$

**= 9 mol NH**_{3 }

**9 moles of NH _{3} are produced from 3.5 mol of H_{2 }and excess N_{2 }**

3 H_{2 }(g) + N_{2 }(g)→ 2 NH_{3}(g)

a. How many moles of NH_{3 }can be produced from 13.5 mol of H_{2 }and excess N_{2 }?

b. How many grams of NH_{3 }can be produced from 4.90 mol of N_{2 }and excess H_{2 }.

c. How many grams of H_{2 }are needed to produce 12.74 g of NH_{3 }?

d. How many molecules (not moles) of NH_{3 }are produced from 1.05×10^{−4} g of H_{2 }?

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