Finding moles and Grams of the equation...?

Question

The complete combustion of octane, C8H18, a component of gasoline, proceeds as follows

2C8H18 + 25O2=============>16CO2 + 18H2O

a) how many moles of O2 are needed to burn 1.50 mol of C8H18?

b) How many grams of O2 are needed to burn 1.50 g of C8H18?

c)Octane has a density of 0.692 g/mL at 20 degrees C. How many grams of O2 are required to burn 1.00 gal of C8H18?

Answer 1

a) 1.50 moles C8H18 x (25 moles O2 / 2 moles C8H18) = 18.8 moles O2

b) 1.50 g C8H18 x (1 mole C8H18 / 114 g C8H18) x (25 moles O2 / 2 moles C8H18) x (32.0 g O2 / 1 mole O2) = 5.26 g O2

c) 1.00 gal C8H18 x (3.79 L / gal) x (1000 mL / L) x (0.692 g C8H18 / mL) x (1 mole C8H18 / 114 g C8H18) x (25 moles O2 / 2 moles C8H18) x (32.0 g O2 / 1 mole O2) = 9200 g O2