Question

Calculate the mass of water produced when 6.48 g of butane reacts with excess oxygen.

Answer 1

C₄H₁₀ + 5.5O2 --------4CO₂ + 5H₂O

If 6.48g of butane react =(6.48/58)= 0.11moles

Then 0.55moles of water are produced from the reaction, which is equivalent to;

$(0.55/1)× 18 =$ 9.9 g