Respuesta :
Answer : The percent yield of the reaction is, 68.2 %
Explanation : Given,
Mass of [tex]SiO_2[/tex] = 90.0 g
Mass of [tex]SiC[/tex] = 41.0 g
Molar mass of [tex]SiO_2[/tex] = 60.08 g/mol
Molar mass of [tex]SiC[/tex] = 40.11 g/mol
First we have to calculate the moles of [tex]SiO_2[/tex]
[tex]\text{Moles of }SiO_2=\frac{\text{Given mass }SiO_2}{\text{Molar mass }SiO_2}[/tex]
[tex]\text{Moles of }SiO_2=\frac{90.0g}{60.08g/mol}=1.498mol[/tex]
Now we have to calculate the moles of [tex]SiC[/tex]
The balanced chemical equation is:
[tex]SiO_2+3C\rightarrow SiC+2CO[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]SiO_2[/tex] react to give 1 mole of [tex]SiC[/tex]
So, 1.498 mole of [tex]HCl[/tex] react to give 1.498 mole of [tex]SiC[/tex]
Now we have to calculate the mass of [tex]SiC[/tex]
[tex]\text{ Mass of }SiC=\text{ Moles of }SiC\times \text{ Molar mass of }SiC[/tex]
[tex]\text{ Mass of }SiC=(1.498moles)\times (40.11g/mole)=60.08g[/tex]
Now we have to calculate the percent yield of the reaction.
[tex]\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield = 41.0 g
Theoretical yield = 60.08 g
Now put all the given values in this formula, we get:
[tex]\text{Percent yield}=\frac{41.0g}{60.08g}\times 100=68.2\%[/tex]
Therefore, the percent yield of the reaction is, 68.2 %
