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%

Question

contestada

Respuesta :

Mergus Mergus · Answer 1 · 2019-11-06T05:20:16+01:00

Answer:

d.57.2%

Explanation:

Given,

Moles of [tex]C_3H_8[/tex] = 6.85 moles

Moles of [tex]O_2[/tex] = 24.9 moles

According to the given reaction:

[tex]C_3H_8+5O_2\rightarrow 3CO_2+4H_2O[/tex]

1 mole of [tex]C_3H_8[/tex] reacts with 5 moles of [tex]O_2[/tex]

6.85 moles of [tex]C_3H_8[/tex] reacts with 5*6.85 moles of [tex]O_2[/tex]

Moles of [tex]O_2[/tex] = 34.25 moles

Available moles of [tex]O_2[/tex] = 24.9 moles

Limiting reagent is the one which is present in small amount. Thus, [tex]O_2[/tex] is limiting reagent.

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

5 moles of [tex]O_2[/tex] produces 3 moles of [tex]CO_2[/tex]

1 mole of [tex]O_2[/tex] produces 3/5 moles of [tex]CO_2[/tex]

24.9 moles of [tex]O_2[/tex] produces (3/5)*24.9 moles of [tex]CO_2[/tex]

Moles of [tex]CO_2[/tex] produced = 14.94 moles

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

[tex]\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100[/tex]

Theoretical yield = 14.94 moles

Given, Experimental yield = 8.55 moles

Applying the values in the above expression as:-

[tex]\%\ yield =\frac{8.55}{14.94}\times 100[/tex]

[tex]\%\ yield =57.2\ \%[/tex]