Answer:
2.5 g
Step-by-step explanation:
The balanced chemical equation for the reaction is:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we can see that for every 1 mole of methane reacted, 2 moles of water are produced. Therefore, we need to first calculate the number of methane and oxygen moles in the reaction mixture.
The molar mass of methane is 16.04 g/mol, so the number of moles of methane present is:
2.41 g / 16.04 g/mol = 0.15 mol
The molar mass of oxygen is 32.00 g/mol, so the number of moles of oxygen present is:
4.6 g / 32.00 g/mol = 0.14 mol
We can see insufficient oxygen to react completely with all of the methane, so oxygen is the limiting reagent. To determine the maximum amount of water that can be produced, we need to use the number of moles of oxygen to calculate the number of moles of water produced and then convert that to a mass using the molar mass of water.
The balanced equation shows that 2 moles of water are produced for every 2 moles of oxygen. Therefore, the number of moles of water produced is:
0.14 mol O2 × (2 mol H2O / 2 mol O2) = 0.14 mol H2O
The molar mass of water is 18.02 g/mol, so the mass of water produced is:
0.14 mol H2O × 18.02 g/mol = 2.53 g
Rounding to 2 significant digits, the maximum mass of water that could be produced is 2.5 g.