# How many milliters of oxygen are required to react completely with 175 mL of C4H10 if the volumes of both gases are measured at the same temperature and pressure? The reaction is 2C4H(g) + 13O2(g) --> 8CO2(g) +10H2O(g)

Mar 24, 2015

You would need $\text{1140 mL}$ of oxygen to react with that much butane.

$\textcolor{red}{2} {C}_{4} {H}_{10 \left(g\right)} + \textcolor{b l u e}{13} {O}_{2 \left(g\right)} \to 8 C {O}_{2 \left(g\right)} + 10 {H}_{2} {O}_{\left(g\right)}$

Notice that you have a $\textcolor{red}{2} : \textcolor{b l u e}{13}$ mole ratio between butane and oxygen, which means that, regardless of how many moles of butane react, you'll always going to need 13/2 times more moles of oxygen for the reaction to take place.

If the volumes of both gases are measured at the same temperature and pressure, you can use the ideal gas law equation to try and figure out what volume of oxygen you'd need to react with that much butane.

So, let's assume temperature is equal to $T$ and pressure is equal to $P$. This means that

$P {V}_{\text{butane") = n_("butane}} \cdot R T$ (1), and

$P {V}_{\text{oxygen") = n_("oxygen}} \cdot R T$ (2)

Here is where the mole ratio that exists between butane and oxygen comes in handy; you know that, regardless of how many moles of butane you have, you need 13/2 times more moles of oxygen.

Mathematically, this can be written as

${n}_{\text{oxygen") = 13/2 * n_("butane}}$

Plug this into equation (2) and divide equation (1) by equation (2) and you'll get

$\frac{\left(1\right)}{\left(2\right)} \implies \left(\cancel{P} {V}_{\text{butane"))/(cancel(P)V_("oxygen")) = (n_("butane") * cancel(RT))/(n_("oxygen}} \cdot \cancel{R T}\right)$

V_("butane")/V_("oxygen") = cancel(n_("butane"))/(13/2 * cancel(n_("butane"))

${V}_{\text{butane")/V_("oxygen") = 2/13 => V_("oxygen") = 13/2 * V_("butane}}$

V_("oxygen") = 13/2 * "175 mL" = "1137.5 mL"

Rounded to three sig figs, the number of sig figs given for 175 mL, the answer will be

${V}_{\text{oxygen") = color(red)("1140 mL}}$