# Question d5e50

May 24, 2016

$\text{4.2 g}$

#### Explanation:

The balanced chemical equation that describes this synthesis reaction looks like this

$\textcolor{red}{3} {\text{Mg"_ ((s)) + "N"_ (2(g)) -> "Mg"_ 3"N}}_{2 \left(s\right)}$

Notice that in order to produce $1$ mole of magnesium nitride, the reaction must consume $\textcolor{red}{3}$ moles of magnesium metal and $1$ mole of nitrogen gas.

You already know how many moles of magnesium metal you have at your disposal, so use the $\textcolor{red}{3} : 1$ mole ratio that exists between the two reactants to determine how many moles of nitrogen gas are needed

0.45color(red)(cancel(color(black)("moles Mg"))) * "1 mole N"_2/(color(red)(3)color(red)(cancel(color(black)("moles Mg")))) = "0.15 moles N"_2#

All you have to do now is use the molar mass of nitrogen gas to determine how many grams of nitrogen gas would contain that many moles

$0.15 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles N"_2))) * "28.01 g"/(1color(red)(cancel(color(black)("mole N"_2)))) = color(green)(|bar(ul(color(white)(a/a)"4.2 g N}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the number of moles of magnesium.