How many grams of O2 are needed to completely react with 5.0 g of butane in a butane lighter?

Question

I was given the balanced equation: 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O. I have done the conversion to find that 5 g of butane is .09 moles of butane. I am unsure where to start when it comes to the oxygen.

Answer 1

Hello, Gabrielle,

The balanced equation tells us that 13 moles of O2 are needed for every 2 moles of butane. If you have 0.09 moles of butane [thanks for doing that calculation], you need to multiply that by the molar ratio (13 moles O2/2 moles butane). So 0.09 x 2/13 =0.56 moles O2. The molar mass of O2 is 32 gram/mole, so multiply that times the moles O2 required to determine the number of grams of O2. I get 17.89 grams, 18 grams with 2 sig figs, although technically it should be 1 sig fig, since the 5 grams of butane is only 1. 20 grams with 1 sig fig, but it hurts to do all that work for such a loose answer. :)

I hope this helps,

Bob