Question

what is the percent yield of H2O if 6.75 grams of H2O are produced in the laboratory from a reaction of 7.73 grams of C3H8 and excess oxygen?C3H8+O2-->>CO2+H20

Answer 1

1) Balance the equation.

C3H8 + 5 O2 => 3 CO2 + 4 H2O

2) Set the equation

`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}}=`
`=\text{ 12,6324 g H2O}`

3) Percentage yield

Expected (from the equation): 12,6324 g of H2O

Actual produced: 6,75 g of H2O

`\text{\% of H2O = }\frac{6,75\text{ g of H2O}}{12,6324\text{ g H2O}}\cdot100\text{ = 53,43\%}`