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=(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* \text{ Molar mass of }SiC`

`\text{ Mass of }SiC=(1.498moles)* (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}}* 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}=(41.0g)/(60.08g)* 100=68.2\%`

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