Question

missed this? watch kcv: limiting reactant, theoretical yield, and percent yield, iwe: limiting reactant and theoretical yield; read section 4.4. you can click on the review link to access the section in your etext. magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. the balanced equation for the reaction is: 2mg(s) + o₂(g) → 2mgo(s) when 13.2 g mg reacts with 13.8 g o₂, 12.7 g mgo is collected. part a part b part c determine the percent yield for the reaction. express your answer as a percent.

Explanation:

Step1: Calculate moles of Mg

$n_{Mg}=\frac{m_{Mg}}{M_{Mg}}$, where $m_{Mg} = 13.2\ g$ and $M_{Mg}=24.31\ g/mol$. So $n_{Mg}=\frac{13.2\ g}{24.31\ g/mol}\approx0.543\ mol$.

Step2: Calculate moles of $O_2$

$n_{O_2}=\frac{m_{O_2}}{M_{O_2}}$, where $m_{O_2} = 13.8\ g$ and $M_{O_2}=32.00\ g/mol$. So $n_{O_2}=\frac{13.8\ g}{32.00\ g/mol}\approx0.431\ mol$.

Step3: Determine the limiting reactant

From the balanced - equation $2Mg(s)+O_2(g)\to2MgO(s)$, the mole ratio of $Mg$ to $O_2$ is $2:1$. For $0.431\ mol$ of $O_2$, the amount of $Mg$ required is $2\times0.431\ mol = 0.862\ mol$. But we have only $0.543\ mol$ of $Mg$. So $Mg$ is the limiting reactant.

Step4: Calculate the theoretical yield of $MgO$

From the balanced - equation, the mole ratio of $Mg$ to $MgO$ is $1:1$. So the moles of $MgO$ produced theoretically, $n_{MgO - theoretical}=n_{Mg}=0.543\ mol$. The molar mass of $MgO$ is $M_{MgO}=24.31\ g/mol + 16.00\ g/mol=40.31\ g/mol$. The theoretical yield of $MgO$, $m_{MgO - theoretical}=n_{MgO - theoretical}\times M_{MgO}=0.543\ mol\times40.31\ g/mol\approx21.9\ g$.

Step5: Calculate the percent yield

Percent yield $=\frac{m_{MgO - actual}}{m_{MgO - theoretical}}\times100\%$. Given $m_{MgO - actual}=12.7\ g$ and $m_{MgO - theoretical}\approx21.9\ g$. Percent yield $=\frac{12.7\ g}{21.9\ g}\times100\%\approx57.9\%$.

Answer:

$57.9\%$