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

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 $Mg + 2HCl\rightarrow MgCl_2+H_2$, 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%