Respuesta :
The percent yield of a reaction that produces 12.5 g of the gas freon cf2cl2 from 32.9g of ccl4 and excess HF is 48.33%
CCl4 + 2HF → CCl2F2 + 2HCl
Mass of CCl4 = 32.9 g
Mass of CCl2F2 (practical yield) = 12.5 g
Molecular mass of CCl4 = 153.82 g / mol
Number of moles of CCl4 = Mass of CCl4 / Molecular mass of CCl4
= 32.9 / 153.82
= 0.2139 mol
Number of moles of CCl4 = Number of moles of CF2Cl2 = 0.2139 mol
Molecular mass of CF2Cl2 = 120.91 g / mol
Mass of CCl2F2 = Number of moles of CF2Cl2 * Molecular mass of CF2Cl2
= 0.2139 * 120.91
Mass of CCl2F2 ( Theoretical yield ) = 25.8626 g
Percentage yield = Practical yield / Theoretical yield * 100
= 12.5 / 25.8626 * 100
Percentage yield = 48.33%
Hence, the percentage yield of CF2Cl2 is 48.33%
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The percent yield of a reaction that produces 12.5 g of the gas from CF2Cl2 from 32.9 g of CCl4 and excess HF is 48.33%.
Chemical equation:
CCl4 + 2HF --- CCl2F2 + 2HCl
Mass of CCl4 = 32.9g
Mass of CCl2F2(practical yield) = 12.5g
Molar mass of CCl4 = 153.82 g/mil
Number of moles of CCl4= given mass / molar mass
= 32.9/153.82
= 0.2139mol
As we know
Number of moles of CCl4 = Number of moles of CF2Cl2 = 0.2139 mol
Molecular mass of CF2Cl2 = 120.91 g/ mol
Mass of CCl2F2= number of moles of CF2Cl2 × molar mass of CF2Cl2= 0.2139× 120.91
Mass of CF2Cl2 (Theoretical yield) = (12.5/25.8626) × 100
Percentage yield= 48.33%
Thus, we concluded that the percentage yield of CF2Cl2 is 48.33%.
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