C₄H₁₀ + [tex] \frac{13}{2} [/tex]O₂ → 4CO₂ + 5H₂O
If mass of butane = 9.69 g
and mole = mass ÷ molar mass
then mole of butane = 9.69 g ÷ ((12 × 4) + (1 × 10))
= 9.69 g ÷ 58
= 0.1671 mol
Mole Ratio of C₄H₁₀ : H₂O is 1 : 5
∴ if mole of butane = 0.1671 mol
then mole of water = (0.1671 mol × 5 )
= 0.8353 mol
Since mass = moles × molar mass
then mass of water produced = 0.8353 mol × ((1 × 2) + (16 × 1))
= 15.036 g
