Question

Consider the reaction of C3H8 with O2 to form CO2 and H2O. If 5.11 g O2 is reacted with excess C3H8 and 3.35 g of CO2 is ultimately isolated, what is the percent yield for the reaction? Percent yield = %

Answer 1

Answer : The percent yield of the reaction is, 79.8 %

Explanation :  Given,

Mass of $O_2$ = 5.11 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $O_2$.

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{5.11g}{32g/mole}=0.159mole$

Now we have to calculate the moles of $CO_2$.

The balanced chemical reaction will be,

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From the balanced reaction, we conclude that

As, 5 moles of $O_2$ react to give 3 moles of $CO_2$

So, 0.159 moles of $O_2$ react to give $\frac{3}{5}\times 0.159=0.0954$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=(0.0954mole)\times (44g/mole)=4.1976g$

The theoretical yield of $CO_2$  = 4.1976 g

The actual yield of $CO_2$  = 3.35 g

Now we have to calculate the percent yield of $CO_2$

$\%\text{ yield of }CO_2=\frac{\text{Actual yield of }CO_2}{\text{Theoretical yield of }CO_2}\times 100=\frac{3.35g}{4.1976g}\times 100=79.8\%$

Therefore, the percent yield of the reaction is, 79.8 %