Answer:
Mass = 17.28 g
Explanation:
Given data:
Mass of propane = 35 g
Volume of oxygen = 27 L
Mass of water produced = ?
Solution:
Chemical equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
First of all we will calculate the number of moles of oxygen at STP.
PV = nRT
n = PV/RT
n = 1 atm×27 L / 0.0821 atm. L/ mol.K × 273 K
n = 27 atm . L / 22.413 atm. L/mol
n = 1.2 mol
Moles of propane:
Number of moles = mass/ molar mass
Number of moles = 35 g/ 44.1 g/mol
Number of moles = 0.8 mol
Now we will compare the moles of water with propane and oxygen.
C₃H₈ : H₂O
1 : 4
0.8 : 4×0.8 = 3.2 mol
O₂ : H₂O
5 : 4
1.2 : 4/5×1.2 = 0.96 mol
Number of moles of water produced by oxygen are less so it will limiting reactant.
Mass of water:
Mass = number of moles × molar mass
Mass = 0.96 mol × 18 g/mol
Mass = 17.28 g
