Answer:
7.676 g of water
Step-by-step explanation:
The combustion of a hydrocarbon yields water and carbon dioxide.
The complete combustion of butane is given by the equation;
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
We are given 4.95 g C₄H₁₀
We need to calculate the mass of water produced.
Step 1: Moles of butane in 4.95 g
Number of moles = Mass ÷ Molar mass
Molar mass = 58.12 g/mol
Number of moles = 4.95 g ÷ 58.12 g/mol
= 0.0852 moles
Step 2: Moles of water
From the equation;
2 moles of butane completely burns in air to produce 10 moles of water
Therefore, 0.0852 moles of butane will produce;
= 0.0852 moles × 10/2
= 0.426 moles
Step 3: Mass of water produced
Mass is given by multiplying number of moles of a compound by its molar mass.
Thus;
Mass of water = 0.426 moles × 18.02 g/mol
= 7.676 g
Therefore, 7.676 g of water are produced when 4.95 g of butane completely burns in excess oxygen.