Respuesta :
Answer : The percent yield of the reaction is, 79.8 %
Explanation : Given,
Mass of [tex]O_2[/tex] = 5.11 g
Molar mass of [tex]O_2[/tex] = 32 g/mole
Molar mass of [tex]CO_2[/tex] = 44 g/mole
First we have to calculate the moles of [tex]O_2[/tex].
[tex]\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{5.11g}{32g/mole}=0.159mole[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex].
The balanced chemical reaction will be,
[tex]C_3H_8+5O_2\rightarrow 3CO_2+4H_2O[/tex]
From the balanced reaction, we conclude that
As, 5 moles of [tex]O_2[/tex] react to give 3 moles of [tex]CO_2[/tex]
So, 0.159 moles of [tex]O_2[/tex] react to give [tex]\frac{3}{5}\times 0.159=0.0954[/tex] moles of [tex]CO_2[/tex]
Now we have to calculate the mass of [tex]CO_2[/tex]
[tex]\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2[/tex]
[tex]\text{Mass of }CO_2=(0.0954mole)\times (44g/mole)=4.1976g[/tex]
The theoretical yield of [tex]CO_2[/tex] = 4.1976 g
The actual yield of [tex]CO_2[/tex] = 3.35 g
Now we have to calculate the percent yield of [tex]CO_2[/tex]
[tex]\%\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\%[/tex]
Therefore, the percent yield of the reaction is, 79.8 %
