# Question e73d1

Jan 23, 2018

$\text{7.0 moles N}$

#### Explanation:

All you have to do here is to identify how many moles of nitrogen are present in $1$ mole of ammonium nitrate, ${\text{NH"_4"NO}}_{3}$.

Now, you can determine how many moles of each element are present per mole of ammonium nitrate by looking at the chemical formula of the compound.

Every subscript added after an element acts as a multiplier. if a subscript follows a pair of parentheses, the subscript is distributed to all the elements present in the parentheses.

Also, don't forget that a subscript of $1$ is not added to the chemical formula. So every time you see an element that is not followed by a subscript, know that the element actually has a subscript of $1$.

So, you can say that $1$ mole of ammonium nitrate will contain

"1 mole NH"_4"NO"_3 => { (1 xx "N"), (4 xx "H"), (1 xx "N"), (3 xx "O") :}#

This means that for every mole of ammonium nitrate, you get $2$ moles of nitrogen. You can thus say that your sample will contain

$3.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NH"_4"NO"_3))) * "2 moles N"/(1color(red)(cancel(color(black)("mole NH"_4"NO"_3)))) = color(darkgreen)(ul(color(black)("7.0 moles N}}}}$

The answer is rounded to two sig figs.