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Need help with % yield please : (

When 48.91 grams of SiC are reacted with 14.048 liters of Cl2, 31.527 grams of SiCl4 are produced at STP. What is the percent yield of this reaction?

Need help with % yield please : ( When 48.91 grams of SiC are reacted with 14.048 liters-example-1
User Rochb
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1 Answer

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To calculate the percent yield of the reaction, we first need to calculate the theoretical yield of the reaction, which we can calculate using stoichiometry:

From the balanced chemical equation: SiC + 2Cl2 -> SiCl4 + C

Molar mass of SiC = 40.10 g/mol
Molar mass of Cl2 = 70.91 g/mol
Molar mass of SiCl4 = 169.9 g/mol

Number of moles of SiC = 48.91 g / 40.10 g/mol = 1.221 mol
Number of moles of Cl2 = (14.048 L) * (1 mol / 22.414 L) = 0.6268 mol (using the ideal gas law at STP)
Using the stoichiometric coefficients, we can see that 1 mol of SiC produces 1 mol of SiCl4, so the theoretical yield of SiCl4 is equal to the number of moles of SiC, which is 1.221 mol.

We can calculate the mass of theoretical yield of SiCl4 using its molar mass:
Mass of SiCl4 (theoretical) = 1.221 mol * 169.9 g/mol = 207.4 g

Now, we can calculate the percent yield:

Percent yield = (actual yield / theoretical yield) * 100%

We are given that the actual yield of SiCl4 is 31.527 grams.

Percent yield = (31.527 g / 207.4 g) * 100% = 15.2%

Therefore, the percent yield of the reaction is approximately 15.2%.
User Kolek
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