Question

C5H12 (l) + 8O2 (g) → 5CO2 (g) + 6 H2O (g)

1. Given 4.3 x 1024 atoms of oxygen (O), how many moles of carbon dioxide (CO2) are present? Answer with 2 sig figs.

Answer 1

By the balanced reaction the proportion of the $O_2$ and $CO_2$ mol number is:

$8n(O_2) = 5n(CO_2)$

By the given details, the mol number of the $O_2$ can be deduced from the following formula:

$n(O_2) = \frac{N(O_2)}{N_A}$ , where the $N_A= 6.022 \times10^{23} p/mol$ Avogadro's number.

$n(O_2) = \frac{4.3 \times 10^{24}}{6.022 \times 10^{22}} = 0.71\times10^2 mol$

Using the first proportion from the balanced equation the mol number of the carbon dioxide is:

$n(CO_2)=\frac{8n(O_2)}{5} = \frac{8\times 0.71\times 10^2}{5} = 1.1 \times 10^2 mol$