Stoichiometry - 3 | Chemical Calculations

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.
n_O2 = 5 x n_C3H8
n_O2 = 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
m_O2 = n_O2 x M_O2 = 1.13 x 32.0 = 36.3 g

3) From the chemical equation: four moles of water are produced for each mole of C₃H₈.
n_H2O = 4 n_C3H8
n_H2O = 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
m_H2O = n_H2O x M_H20 = 9.07 x 10^-1 x 18 = 16.3 g