Calculate the mass of water produced when 4.95 g of butane reacts with excess oxygen.

7.676 g  of water

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.

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

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

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.

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