1) Balance the equation.
C3H8 + 5 O2 => 3 CO2 + 4 H2O
2) Set the equation
[tex]gramsofH2O=\frac{7,73\text{ g C3H8}}{\square}\cdot\frac{1\text{ mol C3H8}}{44,0956\text{ g C3H8}}\cdot\frac{4\text{ mol H2O}}{1\text{ mol C3H8}}\cdot\frac{{}18,01528\text{ g H2O}}{1\text{ mol H2O}}=[/tex][tex]=\text{ 12,6324 g H2O}[/tex]3) Percentage yield
Expected (from the equation): 12,6324 g of H2O
Actual produced: 6,75 g of H2O
[tex]\text{\% of H2O = }\frac{6,75\text{ g of H2O}}{12,6324\text{ g H2O}}\cdot100\text{ = 53,43\%}[/tex]