Answer:
The balanced equation:
C3H8 + 5O2➡️3CO2 + 4H2O
First, let’s convert the masses of C3H8 and O2 to moles by the formula of (mass of substance)/(molar mass of substance).
Molar mass of C3H8 = (3 x 12.01 + 8 x 1.008) = 44.094 g/mol.
Therefore, moles of C3H8 in 45.89 g = (45.89 g/44.094 g/mol) = 1.041 mol.
Molar mass of O2 = (2 x 16.0) = 32.0 g/mol.
Therefore, moles of O2 in 42.63 g = (42.63 g/32.0 g/mol) = 1.332 mol.
From the balanced equation, the mole ratio of C3H8 to O2 is 1:5.
Therefore, for 1.041 mol of C3H8, moles of O2 required for reaction = (5/1) x 1.041 = 5.205 mol.
But the initial amount of O2 given for reaction with C3H8 is 1.332 mol. Now, the reaction requires 5.205 mol of O2. Therefore, the amount of 1.332 mol of O2 is insufficient for reaction with propane and will be used up first, and so O2 is a limiting reactant. This is important because the limiting reactant will dictate the amount of water vapour formed.
Therefore, C3H8 is an excess reactant in the reaction.
From the balanced equation again, the mole ratio of O2 to water is 5:4.
Therefore, for 1.332 mol of O2, moles of water vapour produced = (4/5) x 1.332 = 1.066 mol.
Molar mass of H2O = (2 x 1.008 + 1 x 16.0) = 18.016 g/mol.
Therefore, mass of water vapour produced = (moles of H2O) x (molar mass of H2O) = (1.066 mol x 18.016 g/mol) = 19.21 g.
Hence, (A) the mass of water vapour produced = 19.21 g and (B) the limiting reactant is O2.