Question

1. in the reaction of mg + 2hcl → mgcl₂ + h₂, 22.8 g of magnesium are reacted with an unlimited amount of hydrochloric acid.

a. what is the theoretical yield, in grams, of magnesium chloride?
b. what is the percent yield of magnesium chloride if 71.1 g are measured in lab?

Explanation:

Step1: Calculate moles of magnesium

The molar mass of Mg is approximately 24.31 g/mol. The number of moles of Mg, $n_{Mg}=\frac{m_{Mg}}{M_{Mg}}$, where $m_{Mg} = 22.8$ g and $M_{Mg}=24.31$ g/mol. So, $n_{Mg}=\frac{22.8}{24.31}\approx0.938$ mol.

Step2: Determine mole - ratio of Mg to $MgCl_2$

From the balanced chemical equation $Mg + 2HCl
ightarrow MgCl_2+H_2$, the mole - ratio of $Mg$ to $MgCl_2$ is 1:1. So, the number of moles of $MgCl_2$ produced, $n_{MgCl_2}=n_{Mg} = 0.938$ mol.

Step3: Calculate theoretical yield of $MgCl_2$

The molar mass of $MgCl_2$ is $M_{MgCl_2}=24.31+(2\times35.45)=95.21$ g/mol. The theoretical yield of $MgCl_2$, $m_{theo}=n_{MgCl_2}\times M_{MgCl_2}=0.938\times95.21\approx89.3$ g.

Step4: Calculate percent yield

The percent yield formula is $\text{Percent Yield}=\frac{m_{actual}}{m_{theo}}\times100\%$. Given $m_{actual} = 71.1$ g and $m_{theo}\approx89.3$ g. So, $\text{Percent Yield}=\frac{71.1}{89.3}\times100\%\approx79.6\%$.

Answer:

a. 89.3 g
b. 79.6%