**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 }?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Stoichiometry concept. You can view video lessons to learn Stoichiometry. Or if you need more Stoichiometry practice, you can also practice Stoichiometry practice problems.