Reacting 991 mol of \(\mathrm{SiO}_{2}\) with excess carbon yields 30.0 \(\mathrm{kg}\) of \(\mathrm{SiC.}\) What is the percentage yield? $$\mathrm{SiO}_{2}+3 \mathrm{C} \rightarrow \mathrm{SiC}+2 \mathrm{CO}$$

Short Answer

Expert verified
The percentage yield of the reaction is approximately 75.47%.

Step by step solution

01

Calculate the Theoretical Yield

From the balanced chemical equation, it is seen that 1 mol of SiO2 produces 1 mol of SiC. Therefore, theoretically, 991 mol of SiO2 should produce 991 mol of SiC. The molar mass of SiC is approximately 40.1 g/mol. So, the theoretical yield of SiC in grams would be \(991 \, mol \times 40.1 \, g/mol = 39739.1 \, g\) or 39.739 kg.
02

Calculate the Actual Yield

The actual yield is given in the problem as 30.0 kg, which is equivalent to 30,000 g.
03

Calculate the Percentage Yield

The percentage yield can be calculated using the formula: \[ Percentage \, Yield = \left( \frac{Actual \, Yield}{Theoretical \, Yield} \right) \times 100\%\] Substituting the actual and theoretical yield calculated in previous steps yields \[Percentage \, Yield = \left( \frac{30000 g}{39739.1 g} \right) \times 100\% \approx 75.47\%\]

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