Answer:
Mass of water produced= 1.8 g
Explanation:
Given data:
Mass of water produced = ?
Mass of butane = 1.36 g
Mass of oxygen = excess
Solution:
Chemical equation:
2C₄H₁₀ +13 O₂ → 8CO₂ + 10H₂O
Number of moles of butane:
Number of moles = mass/molar mass
Number of moles = 1.36 g/ 58.12 g/mol
Number of moles = 0.02 mol
Now we will compare the moles of butane with water.
C₄H₁₀ : H₂O
2 : 10
0.02 : 10/2×0.02 = 0.1 mol
Mass of water produced:
Mass = molar mass × molar mass
Mass = 0.1 mol × 18 g/mol
Mass = 1.8 g
