Step-by-step explanation:
The equation fort the combustion reaction of methane is:
CH₄ + 2 O₂ ---> CO₂ + 2 H₂O
60.0 g of CO₂ are produced and we have to find the amount of H₂O that is also produced. First we have to convert the mass of CO₂ into moles using its molar mass.
molar mass of C = 12.01 g/mol
molar mass of O = 16.00 g/mol
molar mass of CO₂ = 1 * 12.01 g/mol + 2 * 16.00 g/mol
molar mass of CO₂ = 44.01 g/mol
moles of CO₂ = 60.0 g * 1 mol/(44.01 g)
moles of CO₂ = 1.36 moles
CH₄ + 2 O₂ ---> CO₂ + 2 H₂O
According to the coefficients of the equation, when 1 mol of CO₂ is produced, 2 moles of H₂O are also produced. We can use that ratio to find the number of moles of H₂O that are produced during our reaction.
1 mol of CO₂ : 2 moles of H₂O molar ratio
moles of H₂O = 1.36 moles of CO₂ * 2 moles of H₂O/(1 mol of CO₂)
moles of H₂O = 2.72 moles
And finally we can convert back to grams the moles of water using its molar mass.
molar mass of H = 1.01 g/mol
molar mass of O = 16.00 g/mol
molar mass of H₂O = 2 * 1.01 g/mol + 1 * 16.00 g/mol
molar mass of H₂O = 18.02 g/mol
mass of H₂O = 2.72 mol * 18.02 g/(1 mol)
mass of H₂O = 49.0 g
Answer: 49.0 g of steam are also produced.