Step-by-step explanation:
A scientist performs the following reaction:
C₃H₈ (g) + 5 O₂ (g) --> 3 CO₂ (g) + 4 H₂O (g)
The mass of propane that is reacting with excee oxygen gas is 355.0 g. First we have to convert that mass into moles using the molar mass of propane.
molar mass of C = 12.0 g/mol
molar mass of H = 1.0 g/mol
molar mass of C₃H₈ = 3 * 12.0 g/mol + 8 * 1.0 g/mol
molar mass of C₃H₈ = 44.0 g/mol
moles of C₃H₈ = 355.0 g * 1 mol/(44.0 g)
moles of C₃H₈ = 8.068 moles
We have to find the mass of water that is produced, but before we find it, we have to find the number of moles of water that are produced.
C₃H₈ (g) + 5 O₂ (g) --> 3 CO₂ (g) + 4 H₂O (g)
According to the coefficients of the balanced equation, 1 mol of C₃H₈ (when reacting with excess oxygen gas) will produce 4 moles of H₂O. Then, the molar ratio between C₃H₈ and H₂O is 1 to 4. We can use that relationship to find the number of moles of water that are produced from 8.068 moles of C₃H₈.
1 mol of C₃H₈ : 4 moles of H₂O molar ratio
moles of H₂O = 8.068 moles of C₃H₈ * 4 moles of H₂O/(1 mol of C₃H₈)
moles of H₂O = 32.27 moles
Finally we can convert those moles of water back to grams using its molar mass.
molar mass of H = 1.0 g/mol
molar mass of O = 16.0 g/mol
molar mass of H₂O = 2 * 1.0 g/mol + 1 * 16.0 g/mol
molar mass of H₂O = 18.0 g/mol
mass of H₂O = 32.27 moles * 18 g/(1 mol)
mass of H₂O = 580.9 g
Answer: 580.9 g of water is likely to form.