If we are burning ethanol, how many g of O2 are required for the combustion of 421 g of ethanol?

Answer 1

Ethanol is a two-carbon alcohol that has the empirical formula
C2H6O. Its mass per mole is therefore 2*12+6+16=46g/mole

The formula for combustion is:

C2H6O + 3O2 -->heat--> 2CO2 + 3H2O

We have 421/46=9.15 moles of ethanol, and we require

9.15*3=27.46moles of O2

Since each mole O2 is 32 grams, in ttal we need:

27.46moles * 32g/mole=878.61 grams O2 required.