If 19.4 g of butane is reacted with 20.5 g of oxygen, what is the limiting reagent?

Question

Use these molar masses:

C = 12.01 g/mol

H = 1.01 g / mole

O = 16.00 g/mole

Molar mass of oxygen: 2 * 16.00 = 32.00 g/mole

Molar mass of butane: (4 * 12.01) + (10 * 1.01) = 58.14 g/mol

moles of O2 = 20.5 g / (32.00 g/mol) = 0.64 mole O2

moles of C4H10 = 19.4 g / (58.14 g/mol) = 0.3336 mole C4H10

Because the butane has less moles than the oxygen, butane is the

limiting reagent and oxygen is the excess.

Answer 1

2 C4H10 & 13 O2 -->  8 CO2  & 10 H2O

1 C4H10 & 6.5 O2 -->  4 CO2  & 5H2O

they need 6.5 times as many moles of oxygen as they have butane

they only added about twice as many moles of oxygen

oxygen is the limiting reagent