Question

Equation: SiO2 + 3C = SiC + 2CO When 90.0 g of silicon dioxide is heated with an excess of carbon, 41.0 g of silicon carbide is produced. What is the percent yield of this reaction? (find the theoretical amount of SiC using stoichiometry, then calculate percent yield)

Answer 1

Answer : The percent yield of the reaction is, 68.2 %

Explanation : Given,

Mass of $SiO_2$ = 90.0 g

Mass of $SiC$ = 41.0 g

Molar mass of $SiO_2$ = 60.08 g/mol

Molar mass of $SiC$ = 40.11 g/mol

First we have to calculate the moles of $SiO_2$

$\text{Moles of }SiO_2=\frac{\text{Given mass }SiO_2}{\text{Molar mass }SiO_2}$

$\text{Moles of }SiO_2=\frac{90.0g}{60.08g/mol}=1.498mol$

Now we have to calculate the moles of $SiC$

The balanced chemical equation is:

$SiO_2+3C\rightarrow SiC+2CO$

From the reaction, we conclude that

As, 1 mole of $SiO_2$ react to give 1 mole of $SiC$

So, 1.498 mole of $HCl$ react to give 1.498 mole of $SiC$

Now we have to calculate the mass of $SiC$

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

$\text{ Mass of }SiC=(1.498moles)\times (40.11g/mole)=60.08g$

Now we have to calculate the percent yield of the reaction.

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 41.0 g

Theoretical yield = 60.08 g

Now put all the given values in this formula, we get:

$\text{Percent yield}=\frac{41.0g}{60.08g}\times 100=68.2\%$

Therefore, the percent yield of the reaction is, 68.2 %