Question

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 97. g of octane is mixed with 150. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

Answer 1

Answer: 61 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles of octane}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{97g}{114g/mol}=0.85moles$

$\text{Number of moles of oxygen}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{150g}{32g/mol}=4.69moles$

The chemical equation for the combustion of octane in oxygen follows the equation:

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

By stoichiometry of the reaction;

25 moles of oxygen react with 2 moles of octane

4.69 moles of oxygen react with=$\frac{2}{25}\times 4.69=0.37$  moles of octane

Thus, oxygen is the limiting reagent as it limits the formation of product and octane is the excess reagent.

25 moles of oxygen produce 18 moles of water

4.69 moles of oxygen produce=$\frac{18}{25}\times 4.69=3.38$  moles of water.

Mass of water produced=$moles\times {\text{Molar mass}}=3.38\times 18g/mol=61g$

The maximum mass of water that could be produced by the chemical reaction is 61 grams.