How many moles of O 2 are required to burn completely 63.5 g of C 6H 6, according to the following equation?

Question

How many moles of O 2 are required to burn completely 63.5 g of C 6H 6, according to the following equation?

2C 6H 6 + 15O 2 ® 12CO 2 + 6H 2O

Answer 1

Hello, Kent,

The balanced equation tells us that we need 15 moles of O2 for every 2 moles of C6H6. That's a molar ratio of 7.5 O2/C6H6.

Calculate the moles that are contained in the 63.5 g of C6H6: The molar mass of C6H6 is 78.0 g/mole. Moles C6H6 = 63.5 g/(78.0 g/mole) or 0.813 moles C6H6.

Multiply by the molar ratio of 7.5 (moles O2)/(mole C6H6) to find the moles of O2 required.

(0.813 moles C6H6)*(7.5 (moles O2)/(mole C6H6)) = 6.10 moles O2