# C3H8 + 5O2 = 3CO2 +4H2O, How much propane is needed to react with 6.2 moles of oxygen?

Feb 13, 2018

$31$ mols ${C}_{3} {H}_{8}$

#### Explanation:

Your equation is ${C}_{3} {H}_{8} + 5 {O}_{2} \to 3 C {O}_{2} + 4 {H}_{2} O$.

Propane is ${C}_{3} {H}_{8}$ and oxygen gas is ${O}_{2}$.

We do this by doing a ratio of the moles of propane to the moles of oxygen gas.

First of all, we must make sure that the equation is balanced. To do so, you have to make sure that the number of atoms on the left are the same as on the right.

There are 3 carbons in the reactants, and 3 carbons in the products. Good.

There are 8 hydrogens in the reactants and 8 hydrogens in the products. Good again.

There are 10 oxygens in the reactants and 10 oxygens in the product. Good again.

This means that the equation is already balanced.

Now we convert the amount of moles of oxygen to the number of moles of propane.

There is one mole of propane (C_3H_8) and five moles of oxygen gas (${O}_{2}$) in the balanced equation.

$6.2$ mols ${O}_{2}$ $\frac{5 m o l s {C}_{3} {H}_{8}}{6 m o l s {O}_{2}}$

$6.2 \cancel{m o l s {O}_{2}}$ $\frac{5 m o l s {C}_{3} {H}_{8}}{1 \cancel{m o l {O}_{2}}}$

Now we are left with the moles of propane:
$31$ mols ${C}_{3} {H}_{8}$.

Hope this helps!