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

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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 
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