The idea here is that you need to determine how many moles of gas remain in the vessel after the reaction is complete.
#"CO"_ ((g)) + "Cl"_ (2(g)) -> "COCl"_ (2(g))#
Notice that the two reactants react in a
Moreover, for every mole of carbon monoxide and chlorine gas that reacts, the reaction produces
Your reaction vessel will contain
#n_"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,
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
You can thus say that
#color(blue)(|bar(ul(color(white)(a/a)V_1/n_1 = V_2/n_2color(white)(a/a)|)))#
Rearrange to solve for
#V_1/n_1 = V_2/n_2 implies V_2 = n_2/n_1 * V_1#
Plug in your values to get
#V_2 = (0.1400 color(red)(cancel(color(black)("moles"))))/(0.2800color(red)(cancel(color(black)("moles")))) * "13.7 L" = color(green)(|bar(ul(color(white)(a/a)color(black)("6.85 L")color(white)(a/a)|)))#
The answer is rounded to three sig figs.