# How many liters of hydrogen gas are produced at STP from the decomposition of "70.0 g" of hydrogen chloride?

Jan 18, 2015

The answer is $\text{21.5 L}$.

So, start with the balanced chemical equation for the decomposition of hydrochloric acid

$2 {\text{HCl" -> "H"_2 + "Cl}}_{2}$

Notice that you have a $\text{2:1}$ mole ratio between $\text{HCl}$ and ${\text{Cl}}_{2}$, which means that every 2 moles of the former will produce 1 mole of the latter. The number of moles of $\text{HCl}$ you have is

"70.0 g" * ("1 mole")/("36.5 g") = 1.92 $\text{moles}$

As a result, the number of ${\text{Cl}}_{2}$ moles is

${n}_{C {l}_{2}} = {n}_{H C l} / 2 = \frac{1.92}{2} = 0.96$ $\text{moles}$

At STP, 1 mole of any ideal gas occupies 22.4 L, which means that the volume of ${\text{Cl}}_{2}$ produced will be

$n = \frac{V}{V} _ \left(\text{molar") => V = n * V_("molar}\right) = 0.96$ $\text{moles} \cdot 22.4$ $\frac{L}{\text{mole}} = 21.5$ $\text{L}$