Acetone and propanal are functional isomers having the same molecular formula C3H6O but different functional groups:
This means that both compounds have the same equation of combustion:
C3H6O + 4O2 = 3CO2 + 3H2O
Using ideal gas law pV=nRT we can find number of moles of carbon dioxide:
"n(CO_2) = \\frac{pV}{RT} = \\frac {101325Pa \\cdot 0.00448m^3}{8.314\\frac{J}{mol \\cdot K}\\cdot 273.15K} = 0.2 mol"
According to stoichiometry of the reaction
"n(H_2O) = n(CO_2) = 0.2 mol"
Thus mass of water produced in this reaction is
"m(H_2O) = n(H_2O)\\cdot M(H_2O) = 0.2 mol \\cdot 18\\frac{g}{mol} = 3.6g"
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