Question

25.3g C3H8 is burned completely and 68.3g of water is measured as a product. Calculate the percent yield.

Answer 1

60.48%

Step-by-step explanation

The combustion of propane is expressed as:

`C_3H_8+5O_2\rightarrow3CO_2+4H_2O`

Determine the moles of water

Moles of water = 68.3/18

Moles of water = 3.79moles

Based on stoichiometry, 1 moles of propane produce 4 moles of water, the moles of propane required will be:

`moles\text{ of C}_3H_8=(3.79)/(4)=0.949moles`

Determine the mass of propane (theoretical yield)

Mass of propane = 0.949 * 44.1

Mass of propane = 41.83grams

Determine the percentage yield

`\begin{gathered} \%yield=(actual)/(theoretical)*100 \\ \%yield=(25.3)/(41.83)*100 \\ \%yield=60.48\% \end{gathered}`

Hence the percent yield is 60.48%