What mass of water is produced from the complete combustion of 2.50×10−3 of methane?
To calculate the mass of water produced from the complete combustion of methane, we need to use the balanced equation for the combustion of methane:
CH₄ + 2O₂ → CO₂ + 2H₂O
From the balanced equation, we can see that 1 mole of methane (CH₄) reacts to produce 2 moles of water (H₂O).
First, we need to determine the number of moles of methane in 2.50×10−3 grams using the molar mass of methane. The molar mass of methane (CH₄) is:
C: 12.01 g/mol
H: 1.008 g/mol x 4 = 4.032 g/mol
Total molar mass of CH₄ = 12.01 + 4.032 = 16.042 g/mol
Now, we can calculate the number of moles of methane using the mass given:
moles of CH₄ = mass of CH₄ / molar mass of CH₄
moles of CH₄ = 2.50×10−3 g / 16.042 g/mol
Next, we use the stoichiometry of the balanced equation to determine the number of moles of water produced:
moles of H₂O = moles of CH₄ x (2 moles of H₂O / 1 mole of CH₄)
Finally, we can calculate the mass of water produced by multiplying the number of moles of water by the molar mass of water:
mass of H₂O = moles of H₂O x molar mass of H₂O
The molar mass of water (H₂O) is:
H: 1.008 g/mol x 2 = 2.016 g/mol
O: 16.00 g/mol
Total molar mass of H₂O = 2.016 + 16.00 = 18.016 g/mol
mass of H₂O = moles of H₂O x 18.016 g/mol