# Question a19b3

Dec 20, 2017

$\text{6.2 g}$

#### Explanation:

The balanced chemical equation that describes this reaction tells you that you need $1$ mole of nitrogen gas to produce $2$ moles of ammonia.

${\text{N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH}}_{3 \left(g\right)}$

So right from the start, you can use the molar mass of nitrogen gas and the molar mass of ammonia to say that every

1 color(red)(cancel(color(black)("mole N"_2))) * "28.0134 g"/(1color(red)(cancel(color(black)("mole N"_2)))) = "28.0134 g"

of nitrogen gas that react, the reaction produces

2 color(red)(cancel(color(black)("moles NH"_3))) * "17.031 g"/(1color(red)(cancel(color(black)("mole NH"_3)))) = "34.062 g"

of ammonia. So if you know that the reaction produced $\text{7.5 g}$ of ammonia, you can use this gram ratio to determine the mass of nitrogen gas that took palce in the reaction, assuming, of course, that the reaction has a 100%# yield.

$7.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g NH"_3))) * "28.0134 g N"_2/(34.062color(red)(cancel(color(black)("g NH"_3)))) = color(darkgreen)(ul(color(black)("6.2 g N}}_{2}}}}$

The answer is rounded to two sig figs.