14.85 g of water
Explanation:
Calculate the mass of water produced when 9.56 g of butane reacts with excess oxygen.
Here we have the full oxidation of butane (C₄H₁₀) which produce carbon dioxide and water.
C₄H₁₀ + (13/2) O₂ → 4 CO₂ + 5 H₂O
number of moles = mass / molecular weight
number of moles of butane = 9.56 / 58 = 0.165 moles
Now, taking in account the chemical reaction, we devise the following reasoning:
if 1 mole of butane produces 5 moles of water
then 0.165 moles of butane produces X moles of water
X = (0.165 × 5) / 1 = 0.825 moles of water
mass = number of moles × molecular weight
mass of water = 0.825 × 18 = 14.85 g
Learn more about:
oxidation of hydrocarbons
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