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