Question

Question #1eeff

Answer 1

`"7 mol H"_2"` will produce `~~5color(white)(.)"mol NH"_3"`.

Explanation:

Balanced Equation

`"N"_2+"3H"_2` `rarr` `"2NH"_3`

Since the amount of `"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 `"H"_2"` by the mole ratio between `"H"_2"` and `"NH"_3"` in the balanced equation. Since we want to end up with `"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