Mass of water produced when 9.69 g of butane reacts with excess oxygen.
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Question:
Mass of water produced when 9.69 g of butane reacts with excess oxygen.
Answers
C₄H₁₀ + [tex] \frac{13}{2} [/tex]O₂ → 4CO₂ + 5H₂OIf mass of butane = 9.69 gand mole = mass ÷ molar massthen mole of butane = 9.69 g ÷ ((12 × 4) + (1 × 10)) = 9.69 g ÷ 58 = 0.1671 molMole Ratio of C₄H₁₀ : H₂O is 1 : 5∴ if mole of butane = 0.1671 molthen mole of water = (0.1671 mol × 5 ) = 0.8353 molSince mass = moles × molar massthen mass of water produced = 0.8353 mol × ((1 × 2) + (16 × 1)) = 15.036 g
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