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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boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey

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