Question

How many grams of oxygen are required to react with 12.0 grams of octane(C8H18) in the combustion of octane in gasoline?

Answer 1

Answer : 42 grams of oxygen are required.

Explanation :

Step 1 : Write balanced chemical equation.

The combustion reaction of octane with oxygen can be written as,

$2 C_{8}H{18} (g) + 25 O_{2} (g) \rightarrow 16 CO_{2} (g) +18 H_{2}O (g)$

Step 2 : Find moles of octane.

The given mass of octane is 12 g

Molar mass of octane is 114.2 g/mol

The moles of octane are calculated as,

$moles =\frac{grams}{MolarMass} =\frac{12g}{114.2g/mol} = 0.105 mol$

We have 0.105 mol octane.

Step 3: Use mole ratio from balanced equation to find moles of O₂

The mole ratio of octane and O₂ is 2 : 25. Let us use this as a conversion factor.

$0.105 mol C_{8}H_{18} \times\frac{25mol O_{2}}{2molC_{8}H_{18}} = 1.31 mol$

We have 1.31 mol O₂.

Step 4 : Convert moles of O₂ to grams.

The molar mass of O₂ is 32 g/mol.

The grams of O₂ can be calculated as,

$1.31 mol O_{2} \times\frac{32g}{mol} = 42 grams$

42 grams of oxygen are required.