Calculate the amount of water produced by the combustion of 44g of propane and 16g of methane.
To calculate the amount of water produced by the combustion of 44g of propane and 16g of methane, we need to first determine the number of moles of each compound.
1. Calculate the number of moles of propane:
- The molar mass of propane (C3H8) is 3(12.01) + 8(1.008) = 44.11 g/mol.
- Moles of propane = Mass of propane / Molar mass of propane
= 44g / 44.11 g/mol
≈ 1 mol
2. Calculate the number of moles of methane:
- The molar mass of methane (CH4) is 12.01 + 4(1.008) = 16.04 g/mol.
- Moles of methane = Mass of methane / Molar mass of methane
= 16g / 16.04 g/mol
≈ 1 mol
Now, let's calculate the number of moles of water produced from each compound using the balanced chemical equations:
1. For propane (C3H8):
C3H8 + 5O2 → 3CO2 + 4H2O
From the balanced equation, we can see that 1 mole of propane produces 4 moles of water.
Moles of water produced from propane = Moles of propane × 4
= 1 mol × 4
= 4 mol
2. For methane (CH4):
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that 1 mole of methane produces 2 moles of water.
Moles of water produced from methane = Moles of methane × 2
= 1 mol × 2
= 2 mol
Finally, we can calculate the total moles of water produced:
Total moles of water = Moles of water from propane + Moles of water from methane
= 4 mol + 2 mol
= 6 mol
To convert moles to grams, we'll use the molar mass of water:
Molar mass of water (H2O) = 2(1.008) + 16.00 = 18.02 g/mol
Therefore, the amount of water produced by the combustion of 44g of propane and 16g of methane is:
Mass of water = Total moles of water × Molar mass of water
= 6 mol × 18.02 g/mol
= 108.12 g
So, the total mass of water produced is approximately 108.12 grams.
To calculate the amount of water produced by the combustion of propane and methane, we need to first determine the balanced equations for their respective combustion reactions.
The combustion of propane (C₃H₈) can be represented by the following balanced equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
The combustion of methane (CH₄) can be represented by the following balanced equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Now, let's calculate the moles of propane and methane separately:
Molar mass of propane (C₃H₈) = 12.01 g/mol (carbon) + 3(1.01 g/mol) (hydrogen) = 44.11 g/mol
Moles of propane = mass of propane (44 g) / molar mass of propane (44.11 g/mol) = 1 mole
Molar mass of methane (CH₄) = 12.01 g/mol + 4(1.01 g/mol) = 16.04 g/mol
Moles of methane = mass of methane (16 g) / molar mass of methane (16.04 g/mol) = 1 mole
Now, let's calculate the moles of water produced:
From the balanced equations, we see that for every mole of propane combusted, 4 moles of water are produced. Similarly, for every mole of methane combusted, 2 moles of water are produced.
Moles of water produced from propane combustion = moles of propane combusted (1 mole) x 4 = 4 moles
Moles of water produced from methane combustion = moles of methane combusted (1 mole) x 2 = 2 moles
To calculate the mass of water produced, we need to know the molar mass of water (H₂O):
Molar mass of water (H₂O) = 2(1.01 g/mol) (hydrogen) + 16.00 g/mol (oxygen) = 18.02 g/mol
Now, let's calculate the mass of water produced:
Mass of water produced from propane combustion = moles of water produced (4 moles) x molar mass of water (18.02 g/mol) = 72.08 g
Mass of water produced from methane combustion = moles of water produced (2 moles) x molar mass of water (18.02 g/mol) = 36.04 g
Therefore, the total amount of water produced by the combustion of 44g of propane and 16g of methane is 72.08 g + 36.04 g = 108.12 g
write the balanced equation
convert grams to moles
find the limiting reagent
see how many moles of it will be used
see how many moles of H2O will result
convert back to grams