Question

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 17. g of octane is mixed with 93.0 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer: 52.8 g of $CO_2$

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of octane

$\text{Number of moles}=\frac{17.0g}{114g/mol}=0.150moles$

b) moles of oxygen

$\text{Number of moles}=\frac{93.0g}{32g/mol}=2.91moles$

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

According to stoichiometry :

2 moles of $C_8H_{18}$ require 25 moles of $O_2$

Thus 0.150 moles of  $C_8H_{18}$ require=$\frac{25}{2}\times 0.150=1.875moles$  of $O_2$

Thus $C_8H_{18}$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

As 2 moles of $C_8H_{18}$ give =  16 moles of $CO_2$

Thus 0.150 moles of $C_8H_{18}$ give =$\frac{16}{2}\times 0.150=1.2moles$  of $CO_2$

Mass of $CO_2=moles\times {\text {Molar mass}}=1.2moles\times 44g/mol=52.8g$

Thus 52.8 g of $CO_2$ will be produced from the given masses of both reactants.