# What is the concentration of 1.12*L of carbon dioxide gas?

Apr 25, 2016

Approx. $0.9 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

$C a C {O}_{3} \left(s\right) + 2 H C l \left(a q\right) \rightarrow C a C {l}_{2} \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right) \uparrow$

We assume that the gas was collected at $298 \cdot K$ and at $1$ $a t m$ pressure. Under these conditions, $1$ $m o l$ of gas, any gas, has a volume of $25.4 \cdot L$

So moles of $C {O}_{2}$ $=$ $\frac{1.12 \cdot \cancel{L}}{25.4 \cdot \cancel{L} \cdot m o {l}^{-} 1}$ $=$ $4.41 \times {10}^{-} 2 \cdot m o l$

Now form the stoichiometric equation, for each mole of gas produced, there were 2 mole of hydrochloric acid.

Thus $\text{Concentration}$ $=$ $\frac{2 \times 4.41 \times {10}^{-} 2 \cdot m o l}{0.100 \cdot L}$ $=$ ??*mol*L^-1

This is a slight over estimate in that the gas collected is saturated with water vapour. This saturated vapour pressure should be subtracted from the pressure.