When 37.1 g of CH4 reacts completely with excess chlorine yielding 52.0 g of HCl, what is the percentage yield, according to CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
a. 84.6
b. 38.6
c. 71.3
d. 28.7
e. 61.4
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To find the percentage yield, we need to compare the actual yield (52.0 g of HCl) to the theoretical yield, which is the maximum amount of product that can be formed based on the balanced equation.
First, we need to calculate the molar mass of CH4 and HCl:
- CH4: 1 carbon atom x 12.01 g/mol + 4 hydrogen atoms x 1.01 g/mol = 16.05 g/mol
- HCl: 1 hydrogen atom x 1.01 g/mol + 1 chlorine atom x 35.45 g/mol = 36.46 g/mol
Next, we need to calculate the moles of CH4 used in the reaction:
- moles of CH4 = mass of CH4 / molar mass of CH4 = 37.1 g / 16.05 g/mol = 2.31 mol
According to the balanced equation, the mole ratio between CH4 and HCl is 1:1. Therefore, the theoretical yield of HCl will also be 2.31 mol.
Next, we need to calculate the actual yield of HCl in grams per mole:
- actual yield of HCl = 52.0 g / 2.31 mol = 22.52 g/mol
Finally, we can calculate the percentage yield by dividing the actual yield by the theoretical yield and multiplying by 100:
- percentage yield = (actual yield / theoretical yield) x 100 = (22.52 g/mol / 36.46 g/mol) x 100 ≈ 61.8
Therefore, the percentage yield is approximately 61.8%.
The closest answer choice is e. 61.4, which is the correct answer.
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