Question

Given the equation below, how many moles of butane (C4H10) must be burned in an excess of O2 to produce 150 g CO2?

2C4H10(g) + 13O2(g) --> 8CO2(g) + 10 H2O(g)

Answer 1

2C4H10(g) + 13O2(g) --> 8CO2(g) + 10 H2O(g)
mols CO2 needed = grams/molar mass = 150/44= 3.41
Convert mols CO2 to mols butane using the coefficients in the balanced equation like this.
3.41 mols CO2 x (2 mols butane/8 mols CO2) = ?

Answer 2

0.85 mol