In a rocket motor fueled with butane, how many kg of liquid oxygen should be provided with each kilogram of butane to provide for complete combustion

Answer

So this is the reaction that happens. 
C4H10 + O2 = CO2 + H2O 

Balanced, it is 
2C4H10 + 8O2 = 8CO2 + 10H2O 

Given 1 kg or 1000 g of butane, use stoichiometry aka factor labeling aka conversions and mole ratios to get to grams of oxygen. 

I'll do an example. Let's form water. Hydrogen is diatomic too. 
2H2 + O2 = 2H2O 

Given 1000 g of Hydrogen, I need to know how many grams of oxygen to use. To convert grams to moles,
I know that 1 mol of H2 equals 2.02 g. Then, for every mole of O2, there are 2 moles of H2. Then converting moles of O2 to grams, I know that one mole of it equals 32 grams. 
[1000 g H2] x [1 mol H2/2.02 g H2] x [1 mol O2/2 mol H2] x [32 g O2/1 mol O2] 

My answer would be  7.9 kg 

Rethink and reinvent