The equilibrium given to you looks like this
$"C"_ ((s)) + "H"_ 2"O"_ ((g)) rightleftharpoons "CO"_ ((g)) + "H"_ (2(g))$
By definition, the reaction quotient,
Two important things to note here
- the reaction quotient can be calculated by using the concentrations of the chemical species that take part in the reaction at any given moment in the course of the reaction
- the concentration of solids and of pure liquids is excluded from the expression of the reaction quotient
In this case, the reaction quotient would take the form -- keep in mind that carbon,
$Q_c = (["CO"] * ["H"_2])/(["H"_2"O"])$
Now, use the volume of the reaction vessel to calculate the concentrations of the chemical species that are of interest here
$color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/(V_"solution"[color(blue)("in liters")])color(white)(a/a)|)))$
You will have
$["H"_2"O"] = "14.0 moles"/"2.25 L" = "6.22 M"$
$["CO"] = "3.60 moles"/"2.25 L" = "1.6 M"$
$["H"_2] = "8.50 moles"/"2.25 L" = "3.78 M"$
Plug these values into the equation that gives you the reaction quotient to find
$Q_c = ("1.6 M" * 3.78 color(red)(cancel(color(black)("M"))))/(6.22color(red)(cancel(color(black)("M")))) = "0.972 M"$
The reaction quotient is usually expressed without added units, so your answer will be
$Q_c = color(green)(|bar(ul(color(white)(a/a)color(black)(0.972)color(white)(a/a)|)))$
The answer is rounded to three .