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(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(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(18g)/(molH_2O)=95.04g`

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