Calculate the mass of H2O that is produced from the combustion of 13.8 mol of C3H8n propane.
C3H8 + 5O2---> 3CO2 + 4H2O
known: 13.8mol C3H8
unknown: mass of H2O
Assuming plenty of oxygen, then
for each mole of propane, one will get 4 moles of water vapor. Figure the moles of H2O, convert that to mass.
Your teacher is just too easy.
To calculate the mass of H2O produced from the combustion of 13.8 mol of C3H8, we need to use the mole ratio between C3H8 and H2O in the balanced chemical equation.
The balanced chemical equation shows that 1 mol of C3H8 produces 4 mol of H2O. Therefore, we can set up a proportion to find the number of moles of H2O produced:
(13.8 mol C3H8) x (4 mol H2O / 1 mol C3H8) = 55.2 mol H2O
Now, to find the mass of H2O, we need to use the molar mass of H2O, which is 18.01528 g/mol. We can multiply the number of moles of H2O by its molar mass to get the mass:
(55.2 mol H2O) x (18.01528 g/mol H2O) = 994.78 g
Therefore, the mass of H2O produced from the combustion of 13.8 mol of C3H8 is approximately 994.78 grams.