Question

The reaction of combustion octane in an engine is known to be only 74.3% efficient. How many grams of water would you expect to form if 90.0 grams of liquid octane are burned in an engine

Answer 1

The reaction of combustion of octane is as follows:

$2C_8H_{18}+25O_2\rightarrow16CO_2+18H_2O$

Convert the given mass of octane to moles using the molar mass of octane:

$90.0g\cdot\frac{mol}{114.23g}=0.79mol$

According to the given reaction, 2 moles of octane produce 18 moles of water. Use this ratio to find how many moles of water are produced from 0.79 moles of octane:

$0.79molC_8H_{18}\cdot\frac{18molH_2O}{2molC_8H_{18}}=7.11molH_2O$

Multiply this amount times the efficiency of the reaction to find the actual amount of water produced:

$7.11molH_2O\cdot74.3\%=5.28molH_2O$

Convert this amount to moles using the molar mass of water:

$5.28molH_2O\cdot\frac{18g}{molH_2O}=95.04g$

It means that the mass of water produced is 95.04g.