# Question #0cd1c

May 30, 2016

$\text{6.85 L}$

#### Explanation:

The idea here is that you need to determine how many moles of gas remain in the vessel after the reaction is complete.

Carbon monoxide, $\text{CO}$, and chlorine gas, ${\text{CL}}_{2}$, will react to form phosgene, ${\text{COCl}}_{2}$, according to the balanced chemical equation

${\text{CO"_ ((g)) + "Cl"_ (2(g)) -> "COCl}}_{2 \left(g\right)}$

Notice that the two reactants react in a $1 : 1$ mole ratio. This tells you that the reaction will always consume equal numbers of moles of carbon monoxide and chlorine gas.

Moreover, for every mole of carbon monoxide and chlorine gas that reacts, the reaction produces $1$ mole of phosgene.

Your reaction vessel will contain $0.1400$ moles of carbon monoxide and $0.1400$ moles of chlorine gas. The total number of moles present in the vessel before the reaction takes place will be

${n}_{\text{total" = "0.1400 moles" + "0.1400 moles" = "0.2800 moles}}$

Now, you have equal numbers of moles of each reactant, which means that both carbon monoxide and chlorine gas will be completely consumed by the reaction.

As a result, $1.400$ moles of phosgene will be produced. The number of moles of gas present in the vessel will thus go from $0.2800$ moles to $0.1400$ moles.

As you know, when pressure and temperature are kept constant, the number of moles of gas and the volume they occupy have a direct relationship $\to$ this is known as Avogadro's Law.

You can thus say that

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {V}_{1} / {n}_{1} = {V}_{2} / {n}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

${n}_{1}$, ${V}_{1}$ - the initial number of moles of gas and the volume they occupy
${n}_{2}$, ${V}_{2}$ - the final number of moles of gas and the volume they occupy

Rearrange to solve for ${V}_{2}$, the volume of the gas after the reaction is complete

${V}_{1} / {n}_{1} = {V}_{2} / {n}_{2} \implies {V}_{2} = {n}_{2} / {n}_{1} \cdot {V}_{1}$

Plug in your values to get

${V}_{2} = \left(0.1400 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(0.2800color(red)(cancel(color(black)("moles")))) * "13.7 L" = color(green)(|bar(ul(color(white)(a/a)color(black)("6.85 L}} \textcolor{w h i t e}{\frac{a}{a}} |}}\right)$

The answer is rounded to three sig figs.