# How can I find the stoichiometry of a reaction from the volumes of the reactants and products given?

I have volume $V$ of gas $\ce{A}$ and an equal volume $V$ of gas $\ce{B}$ at room temperature and pressure. I mix, ignite, and find I have a gas (mixture of $\ce{A}$ and product $\ce{C}$) at room pressure, $325~ \mathrm{^\circ C}$, and total volume $3V$. What was the stoichiometry of the reaction?
\begin{align} \text{a)}&&\ce{2A + B &-> C}\\ \text{b)}&&\ce{A + 2B &-> 2C}\\ \text{c)}&&\ce{2A + B &-> 2C}\\ \text{d)}&&\ce{A – 2B &-> C}\\ \end{align}

The answer is $\text{b)}$ but I can't seem to get the logic.
$V$ vol of $\ce{A}$ plus $V$ vol of $\ce{B}$ reacts to give $2V$ vol of $\ce{C}$. How did we find stoichiometry from this? What about some unreacted $\ce{A}$ ?

In a problem like this I think that the best way to answer the question is to set up the equations and try all the answers. However, in this case we can infer from the text that $B$ need to be the limiting reacting, as all of it reacts but some of $A$ is left unreacted. The only answers providing this are answer b and answer d. However, the latter contains a minus sign which doesn't makes sense chemically, so let's focus on the former.

We can compute the moles involved in the reaction using the ideal gas equation $PV = nRT$. Let's assume the volume is $1 \mathrm{L}$, we can calculate the amount of moles for both $A$ and $B$ (as they have the same volume and are present in the same conditions):

$$n = \dfrac{PV}{RT} = 40.88 \mathrm{mmol}$$

If the stoichiometry is the one proposed in answer b ($\ce{A + 2B -> 2C}$) then the whole quantitity of $B$ will react with half the quantity of $A$ to give the same amount of moles of compound $C$, which will be $n = 40.88 \mathrm{mmol}$. Half of $A$ will be left unreacted, which is $n = 20.44 \mathrm{mmol}$. Using the aforementioned equation we can now compute their volumes:

$$V = \dfrac{nRT}{P}$$

Which is $V_a = 1 \mathrm{L}$ and $V_c = 2 \mathrm{L}$, which in total is $V_{tot} = 3 \mathrm{L}$, exactly three times more than the original volume.

• If you're referring to the different mole value, it's just a typo. It's $n = 40.88$ in the product too. Nov 27 '14 at 16:17
• * How did you know $n=40.88$ in the product too ? * Nov 27 '14 at 16:24
• Also there we started with V vol of A and V vol of B, so how can $final volume = 3V$ $>$ $2V$ ? Nov 27 '14 at 16:38
• It's basic stoichiometry: the stoichiometric coefficients can also be considered as the number of moles reacting. In addition, the volume changes also because the reaction conditions (the temperature) has increased. Nov 27 '14 at 16:45
• If we calculate $V(b)$ = $V$(a)$= 1 L , with each n=$40.88$And$V(c)$=$2$L$. So unless whole of A reacts with B, how can we get 2L of C ? Nov 27 '14 at 16:56