Question

Mass of water produced when 9.69 g of butane reacts with excess oxygen.

Answer 1

C₄H₁₀ +
`(13)/(2)`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