Question

4. Magnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for each trial.

Answer 1

Answer: The average percent yield of MgO is 98.59 %.

Explanation:

The chemical equation follows:

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

• For Trial 1:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of Mg = 0.411 g

Molar mass of Mg = 24.3 g/mol

Plugging values in equation 1:

$\text{Moles of Mg}=\frac{0.411g}{24.3g/mol}=0.0170 mol$

By the stoichiometry of the reaction:

If 2 moles of Mg produces 2 moles of MgO

So, 0.0170 moles of Mg will produce = $\frac{2}{2}\times 0.0170=0.0170mol$ of MgO

We know, molar mass of $MgO$ = 40.3 g/mol

Putting values in equation 1, we get:

$\text{Mass of }MgO=(0.0170mol\times 40.3g/mol)=0.685g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Actual value of the product = 0.675 g

Theoretical value of the product = 0.685 g

Plugging values in equation 1:

$\% \text{yield}=\frac{0.675g}{0.685g}\times 100\\\\\% \text{yield}=98.54\%$

Hence, the % yield of the product is 98.54 %

• For Trial 2:

Given mass of Mg = 0.266 g

Molar mass of Mg = 24.3 g/mol

Plugging values in equation 1:

$\text{Moles of Mg}=\frac{0.266g}{24.3g/mol}=0.011 mol$

By the stoichiometry of the reaction:

If 2 moles of Mg produces 2 moles of MgO

So, 0.011 moles of Mg will produce = $\frac{2}{2}\times 0.011=0.011mol$ of MgO

We know, molar mass of  = 40.3 g/mol

Putting values in equation 1, we get:

$\text{Mass of }MgO=(0.011mol\times 40.3g/mol)=0.443g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Actual value of the product = 0.437 g

Theoretical value of the product = 0.443 g

Plugging values in equation 1:

$\% \text{yield}=\frac{0.437g}{0.443g}\times 100\\\\\% \text{yield}=98.64\%$

Hence, the % yield of the product is 98.64 %

Average of a measurement is calculated by given formula:

$Average=\frac{M_1+M_2}{2}$

where,

$M_1$ = percentage yield for Trial 1 = 98.54 %

$M_2$ = percentage yield for Trial 2 = 98.64 %

Putting values in above equation, we get:

$Average=\frac{98.54+98.64}{2}=98.59\%$

Hence, the average percent yield of MgO is 98.59 %.