Calculate the mass of water formed by the complete reaction of 572 g of propane (C3H8)
with excess oxygen. (Write a balanced equation.)
Answer in units of g.
Same question as above. Stoichiometry problem?
To calculate the mass of water formed by the complete reaction of propane (C3H8) with excess oxygen, we need to first write a balanced equation for the reaction between propane and oxygen. The balanced equation for the complete combustion of propane is:
C3H8 + 5O2 → 3CO2 + 4H2O
From the balanced equation, we can see that for every 1 mole of propane (C3H8), 4 moles of water (H2O) are formed.
Now let's calculate the number of moles of propane (C3H8) in 572 g. To do this, we need to use the molar mass of propane, which is:
Molar mass of C3H8 = (3 * 12.01 g/mol) + (8 * 1.01 g/mol) = 44.11 g/mol
Number of moles of C3H8 = Given mass / Molar mass
Number of moles of C3H8 = 572 g / 44.11 g/mol = 12.98 mol
Therefore, we can conclude that 12.98 moles of propane (C3H8) are reacted.
Since the balanced equation shows that for every 1 mole of propane, 4 moles of water are formed, we can calculate the number of moles of water formed by multiplying the number of moles of propane by the mole ratio:
Number of moles of H2O = 12.98 mol * 4 mol H2O / 1 mol C3H8 = 51.92 mol H2O
Finally, to calculate the mass of water formed, we need to use the molar mass of water, which is:
Molar mass of H2O = (2 * 1.01 g/mol) + (1 * 16.00 g/mol) = 18.02 g/mol
Mass of H2O = Number of moles of H2O * Molar mass of H2O
Mass of H2O = 51.92 mol * 18.02 g/mol = 935.2 g
Therefore, the mass of water formed by the complete reaction of 572 g of propane with excess oxygen is 935.2 g.