Determine how many grams of co2 are produced by burning 4.37 g of c4h10.

Combustion of hydrocarbons is when  C and H containing compounds are burnt in O₂
the balanced chemical reaction for combustion of C₄H₁₀ (butane) is as follows;
2C₄H₁₀ + 13O₂ ---> 8CO₂ + 10H₂O
the stoichiometry of C₄H₁₀ to CO₂ is 2;4, simplified ratio is 1:2
this means that for every 1 mole of butane used up,4 moles of CO₂ are formed
molar mass of butane - (12 g/mol *4) + (1 g/mol * 10) = 58 g/mol
58 g of butane      -  1 mol 
Therefore 4.37 g of butane - 1/58 g/mol * 4.37g = 0.075 mol
1 mol of butane forms --> 4 mol of CO₂
Therefore 0.075 mol of butane forms = 4 x 0.075 mol = 0.3 mol of CO₂
molar mass of CO₂ = 44 g/mol
mass of CO₂ formed = 0.3 mol * 44 g/mol = 13.2 g of CO₂ is formed