In the reaction
C3H8 + 5 O2 "\\rightarrow" 3 CO2 + 4 H2O
1 mol of propane produces 4 mol of water:
"n(C_3H_8) = \\frac{n(H_2O)}{4}"
"n(H_2O) = 4\u00b7n(C_3H_8)" .
The number of the moles of propane is the ratio of its mass "m" to its molar mass "M" (44.10 g/mol):
"n(C_3H_8) = \\frac{m}{M} = \\frac{6.3(g)}{44.10 (g\/mol)} = 0.143 (mol)" .
Making use of the same relation to calculate the mass of water produced ("M = 18.02" g/mol):
"m = nM = 4\u00b70.143\u00b718.02 = 10.3" g.
Answer: 10.3 g of H2O is produced from the reaction of 6.3 g of propane.
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