Calculate the mass of water produced when 6.97 g of butane reacts with excess oxygen

The balanced reaction equation for the combustion of butane is as follows;

C₄H₁₀ + 13/2O₂ - - - > 4CO₂ + 5H₂O

the limiting reactant in this reaction is C₄H₁₀ This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.

stoichiometry of C₄H₁₀ to H₂O is 1:5

mass of butane used - 6.97 g

number of moles - 6.97 g / 58 g/mol = 0.12 mol

then the number of water moles produced - 0.12 mol x 5 = 0.6 mol

Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g