Question

What volume of O2 is required to completely burn 15cm3 of pentane? Equation - C5H12 + O2 - CO2 + H2O.​

Answer 1

Answer:

The volume of $\rm O_2$ required would be $120\; \rm cm^{3}$, assuming that both the pentane and the $\rm O_2\!$ in this question are ideal gases and are under the same temperature and pressure.

Explanation:

Balance the equation for the reaction:

$\rm ?\; C_{5}H_{12} + ?\; O_2 \to ?\; CO_{2} + ?\; H_{2}O$.

Start by setting the coefficient of the molecule with the largest number of atoms to $1$.

In the combustion of alkanes (including pentane,) consider setting the coefficient of the alkane to $1\!$.

$\rm 1\; C_{5}H_{12} + ?\; O_2 \to ?\; CO_{2} + ?\; H_{2}O$.

Number of carbon atoms among the reactants: $5$.

Number of hydrogen atoms among the products: $12$.

By the conservation of atoms, there would need to be the same number of carbon and hydrogen atoms (along with the oxygen atoms) among the products.

Hence, the coefficient of $\rm CO_2$ would be $5$ while the coefficient of $\rm H_2O$ would be $12 / 2 = 6$.

$\rm 1\; C_{5}H_{12} + ?\; O_2 \to 5\; CO_{2} + 6\; H_{2}O$.

There would be $5 \times 2 + 6 \times 1 = 16$ oxygen atoms among the products. Also by the conservation of atoms, there would be the same number of oxygen atoms among the reactants.

Hence, the coefficient of $\rm O_2$ would be $16 / 2 = 8$.

$\rm 1\; C_{5}H_{12} + 8\; O_2 \to 5\; CO_{2} + 6\; H_{2}O$.

The ratio between the coefficient of $\rm O_2$ and $\rm C_{5}H_{12}$ in the balanced equation is:

$\displaystyle \frac{n({\rm O_2})}{n({\rm C_{5}H_{12}})} = \frac{8}{1} = 8$.

In other words, it would take eight $\rm O_2$ molecules to react with one $\rm C_{5}H_{12}$ molecule.

Assume that both $\rm O_2$ and $\rm C_{5}H_{12}$ are ideal gases. Under the same temperature and pressure, the volume of the two gases would be proportional to the number of molecules in each gas:

$\displaystyle \frac{V({\rm O_2})}{V({\rm C_{5}H_{12}})} = \frac{n({\rm O_2})}{n({\rm C_{5}H_{12}})} = \frac{8}{1} = 8$.

In other words, it would take $8\; \rm cm^{3}$ of $\rm O_2$ to react with $1\; \rm cm^{3}$ of $\rm C_{5}H_{12}$ under these assumptions. It would then take $8 \times 15\; \rm cm^{3} = 120\; \rm cm^{3}$ of $\rm O_2\!$ to react with $15\; \rm cm^{3}\!$ of $\rm C_{5}H_{12}\!$.