# Question 1eeff

Apr 27, 2017

$\text{7 mol H"_2}$ will produce $\approx 5 \textcolor{w h i t e}{.} \text{mol NH"_3}$.

#### Explanation:

Balanced Equation

${\text{N"_2+"3H}}_{2}$$\rightarrow$${\text{2NH}}_{3}$

Since the amount of $\text{N"_2}$ was not stated, I am assuming that it is in excess, which means we don't need to worry about it.

All we have to do is multiply the given moles of $\text{H"_2}$ by the mole ratio between $\text{H"_2}$ and $\text{NH"_3}$ in the balanced equation. Since we want to end up with $\text{NH"_3}$, we put it in the numerator.

7color(red)cancel(color(black)("mol H"_2))xxoverbrace(2"mol NH"_3)^color(red)("mole ratio")/(3color(red)cancel(color(black)("mol H"_2)))="5 mol NH"_3"# rounded to one significant figure