# How do you do gas stoichiometry problems at STP?

Mar 19, 2014

Gas stoichiometry problems relate the number of moles of reactants and products, so use the ideal gas law at standard temperature and pressure (273.15 K and 1.00 atm) to solve for the number of moles of gas at a specified volume, V:

$n = \frac{P V}{R T}$

If V is given in liters, then use the gas constant

$R = 0.082054 \frac{L - a t m}{m o l - K}$

Mar 19, 2014

At STP (Standard Temperature and Pressure 273 K and 1 atm) we are able to use Avogadro's number of 22.4 L per mole of gas or
22.4 L/mole as our conversion factor in the stoichiometry.

EXAMPLE

How many liters of hydrogen gas are necessary to react with 10 grams of nitrogen to produce ammonia at standard temperature and pressure?

We begin with a balanced chemical equation

${N}_{2} + 3 {H}_{2} \to 2 N {H}_{3}$

Now using the given value of 10 grams of nitrogen we will convert using stoichiometry

$g r a m s {N}_{2} \to m o l {N}_{2} \to m o l {H}_{2} \to L i t e r s {H}_{2}$

$10 g {N}_{2} x \frac{1 m o l {N}_{2}}{28 g {N}_{2}} = 0.357 m o l {N}_{2}$
(remember ${N}_{2}$ , 2 x 14 g = 28 g)

Use the mole ratio from the balanced chemical equation to convert moles of ${N}_{2}$ to moles of ${H}_{2}$

$0.357 m o l {N}_{2} x \frac{3 m o l {N}_{2}}{1 m o l {N}_{2}} = 1.071 m o l {H}_{2}$

Now use Avogadro's number 22.4 L/mol to convert moles of ${H}_{2}$ to Liters of ${H}_{2}$.

$1.071 m o l {H}_{2} x \frac{22.4 L {H}_{2}}{1 m o l {H}_{2}} = 23.99 L {H}_{2}$

Remember, we can only use the 22.4 L/mole value if the reaction is taking place at Standard Temperature and Pressure (STP = 0 C and
1 atm).

Here's a video that might be helpful: