Final answer:
In the complete combustion of methane, 12 dm³ of oxygen would produce 9.65 grams of water, based on stoichiometry and the molar volume of a gas at STP.
Step-by-step explanation:
The student asked about the mass of water produced from the complete combustion of methane where 12 dm³ of oxygen was used. This question pertains to stoichiometry, which is a concept in chemistry that involves the calculation of reactants and products in chemical reactions.
Chemical Equation and Mole Ratio
First, we need the balanced chemical equation for the combustion of methane (CH₄):
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
From the equation, 1 mole of methane reacts with 2 moles of oxygen to produce 2 moles of water.
Finding the Mass of Water Produced
Using the molar volume of a gas at STP (22.4 dm³/mol), we can calculate the moles of oxygen used:
12 dm³ O₂ x (1 mol O₂ / 22.4 dm³) = 0.536 moles O₂
From the stoichiometry of the reaction, 2 moles of water are produced for every 1 mole of methane. Therefore, the moles of water produced are twice the moles of oxygen used:
0.536 moles O₂ x (2 moles H₂O / 2 moles O₂) = 0.536 moles H₂O
The molar mass of water is approximately 18.015 g/mol, so the mass of water produced is:
0.536 moles H₂O x 18.015 g/mol = 9.65 g
Thus, 9.65 grams of water would have been produced from the complete combustion of methane when 12 dm³ of oxygen were used.