Question

Consider the following balanced equation: 5O2(g) + C3H8(g) → 3CO2(g) + 4H2O(l) If 24.9 moles of O2(g) and 6.85 moles of C3H8(g) are allowed to react to produce 8.55 moles of CO2(g), what is the percent yield of the reaction?
a.45.8%
b.97.7%
c.37.1%
d.57.2%
e.61.8%

Answer 1

Answer:

d.57.2%

Explanation:

Given,

Moles of $C_3H_8$ = 6.85 moles

Moles of $O_2$ = 24.9 moles

According to the given reaction:

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

1 mole of $C_3H_8$ reacts with 5 moles of $O_2$

6.85 moles of $C_3H_8$ reacts with 5*6.85 moles of $O_2$

Moles of $O_2$ = 34.25 moles

Available moles of $O_2$ = 24.9 moles

Limiting reagent is the one which is present in small amount. Thus, $O_2$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

5  moles of $O_2$ produces 3 moles of $CO_2$

1 mole of $O_2$ produces 3/5 moles of $CO_2$

24.9 moles of $O_2$ produces (3/5)*24.9 moles of $CO_2$

Moles of $CO_2$ produced = 14.94 moles

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

$\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100$

Theoretical yield = 14.94 moles

Given, Experimental yield = 8.55 moles

Applying the values in the above expression as:-

$\%\ yield =\frac{8.55}{14.94}\times 100$

$\%\ yield =57.2\ \%$