Question

wintergreen oil is produced in the following scheme from 6.84g of ch₃oh: c₇h₆o₃ + ch₃oh → c₈h₈o₃ + h₂o. if 12.5g of c₈h₈o₃ are actually produced and the reaction proceeds to 38.5% yield, what is the theoretical yield, in grams? 9.63g 4.81g 64.9g 32.5g multiple choice 10 points. determine how many grams of ch₂cl₂ are actually produced when 100.0g of ch₄ react with enough cl₂ according to the following reaction, which proceeds to 45.00% yield: ch₄ + 2cl₂ → ch₂cl₂ + 2hcl. 531.3g 239.1g 100.0g 45.0g

Explanation:

Step1: Recall the percent - yield formula

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%$. We can re - arrange it to solve for the theoretical yield: Theoretical yield=$\frac{\text{Actual yield}}{\text{Percent yield}}\times100$.

Step2: Calculate the theoretical yield for the first reaction

Given actual yield = 12.5 g and percent yield = 38.5%. Substitute the values into the formula: Theoretical yield=$\frac{12.5}{38.5}\times100\approx32.5$ g.

Step3: Calculate the theoretical yield for the second reaction

First, find the molar mass of $CH_4$ and $CH_2Cl_2$.
The molar mass of $CH_4$ ($M_{CH_4}$) = 12.01+4×1.01 = 16.05 g/mol.
The molar mass of $CH_2Cl_2$ ($M_{CH_2Cl_2}$) = 12.01 + 2×1.01+2×35.45=84.93 g/mol.
The number of moles of $CH_4$ ($n_{CH_4}$) in 100.0 g is $n_{CH_4}=\frac{100.0}{16.05}\approx6.23$ mol.
From the balanced equation $CH_4 + 2Cl_2
ightarrow CH_2Cl_2+2HCl$, the mole ratio of $CH_4$ to $CH_2Cl_2$ is 1:1. So the theoretical number of moles of $CH_2Cl_2$ is also 6.23 mol.
The theoretical mass of $CH_2Cl_2$ ($m_{theoretical}$) = $n\times M_{CH_2Cl_2}=6.23\times84.93\approx531.3$ g.
Since the percent yield is 45.00%, the actual yield of $CH_2Cl_2$ is $m_{actual}=\frac{45.00}{100}\times531.3 = 239.1$ g.

Answer:

1. D. 32.5g
2. B. 239.1g