# Ammonia gas decomposes according to the equation: 2NH_3(g) -> N_2(g) + 3H_2(g). If 15.0 L of nitrogen is formed at STP, how many liters of hydrogen will be produced (also measured at STP)?

Jul 20, 2016

$\text{45.0 L}$

#### Explanation:

The thing to remember about reactions that involve gases held under the same conditions for pressure and temperature is that the mole ratio that exists between the gaseous compounds in the balanced chemical equation is equivalent to a volume ratio.

The balanced chemical equation that describes your reaction looks like this

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

Notice that for every $2$ moles of ammonia that take part in the reaction, you get $1$ mole of nitrogen gas and $\textcolor{red}{3}$ moles of hydrogen gas.

When all three gases are kept under the same conditions for pressure and temperature, these mole ratios become equivalent to volume ratios. More specifically, you know that nitrogen gas and hydrogen gas have a $1 : \textcolor{red}{3}$ mole ratio.

When both gases are kept under STP conditions, you will have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{n}_{{\text{N"_ 2)/n_ ("H"_ 2) = V_ ("N"_ 2)/V_ ("H}}_{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In this case, you have a $1 : \textcolor{red}{3}$ volume ratio between nitrogen gas and hydrogen gas. This means that you get

$\frac{1}{\textcolor{red}{3}} = {V}_{{\text{N"_ 2)/V_ ("H"_ 2) implies V_ ("H"_ 2) = color(red)(3) xx V_("N}}_{2}}$

Plug in your values to find

V_("H"_2) = color(red)(3) xx "15.0 L" = color(green)(|bar(ul(color(white)(a/a)color(black)("45.0 L")color(white)(a/a)|)))

The answer is rounded to three sig figs.