Stoichiometry - 3 | Chemical Calculations

1) Write the equation of the combustion of propane (C3H8).

2) How many grams of O2 are required to burn 10.0 g of propane?

3) How many grams of water are produced?

1) C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (l)

2) From the chemical equation: five moles of O2 are required to burn one mole of C3H8.
nO2 = 5 x nC3H8
nO2 = 5 x $\frac{{\mathrm{m}}_{\mathrm{C}3\mathrm{H}8}}{{\mathrm{M}}_{\mathrm{C}3\mathrm{H}8}}$ = 5 x $\frac{10.0}{44.1}$ = 1.13 mol
mO2 = nO2 x MO2 = 1.13 x 32.0 = 36.3 g

3) From the chemical equation: four moles of water are produced for each mole of C3H8.
nH2O = 4 nC3H8
nH2O = 4 x $\frac{{\mathrm{m}}_{\mathrm{C}3\mathrm{H}8}}{{\mathrm{M}}_{\mathrm{C}3\mathrm{H}8}}$ = 4 x $\frac{10.0}{44.1}$ = 9.07 x 10-1 mol
mH2O = nH2O x MH20 = 9.07 x 10-1 x 18 = 16.3 g