# Given the chemical equation "N"_2("g") + "3H"_2("g")"rarr"2NH"_3("g")", how many moles of "NH"_3" can be produced if "10. mol H"_2" reacts completely with nitrogen?

Mar 10, 2015

You will get 6.7 mol of NH₃.

The chemical equation is N₂ + 3H₂ → 2NH₃.

This tells you that 3 mol of H₂ give 2 mol of NH₃. So,

${\text{10. mol H"_2 × ("2 mol NH"_3)/("3 mol H"_2) = "6.7 mol H}}_{2}$

Mar 10, 2015

$\text{10. mol H"_2}$ will produce $\text{6.7 mol NH"_3}$ when it reacts completely with nitrogen.

"N"_2("g") + $\text{3H"_2("g")}$ $\rightarrow$ $\text{2NH"_3("g")}$

From the balanced equation, we can see that the mole ratio of $\text{H"_2}$ to $\text{NH"_3}$ is 3:2. We can write this mole ratio as two conversion factors, one of which will be used to solve the problem.

Two Mole Ratio Conversion Factors

$\text{3 mol H2"/"2 mol NH3}$ and $\text{2 mol NH3"/"3 mol H2}$

We want to know how many moles of ammonia can be produced from 10. moles of hydrogen. We need to multiply 10. moles ${\text{H}}_{2}$ times the mole ratio conversion factor that cancels moles $\text{H"_2}$ and leaves moles ${\text{NH}}_{3}$.

${\text{10. mol H}}_{2}$ x $\text{2 mol NH3"/"3 mol H2}$ = $\text{6.7 mol NH"_3}$