Answer :
Answer: The average percent yield of MgO is 98.59 %.
Explanation:
The chemical equation follows:
[tex]2Mg(s)+O_2(g)\rightarrow 2MgO(s)[/tex]
- 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:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of Mg = 0.411 g
Molar mass of Mg = 24.3 g/mol
Plugging values in equation 1:
[tex]\text{Moles of Mg}=\frac{0.411g}{24.3g/mol}=0.0170 mol[/tex]
By the stoichiometry of the reaction:
If 2 moles of Mg produces 2 moles of MgO
So, 0.0170 moles of Mg will produce = [tex]\frac{2}{2}\times 0.0170=0.0170mol[/tex] of MgO
We know, molar mass of [tex]MgO[/tex] = 40.3 g/mol
Putting values in equation 1, we get:
[tex]\text{Mass of }MgO=(0.0170mol\times 40.3g/mol)=0.685g[/tex]
The percent yield of a reaction is calculated by using an equation:
[tex]\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100[/tex] ......(2)
Given values:
Actual value of the product = 0.675 g
Theoretical value of the product = 0.685 g
Plugging values in equation 1:
[tex]\% \text{yield}=\frac{0.675g}{0.685g}\times 100\\\\\% \text{yield}=98.54\%[/tex]
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:
[tex]\text{Moles of Mg}=\frac{0.266g}{24.3g/mol}=0.011 mol[/tex]
By the stoichiometry of the reaction:
If 2 moles of Mg produces 2 moles of MgO
So, 0.011 moles of Mg will produce = [tex]\frac{2}{2}\times 0.011=0.011mol[/tex] of MgO
We know, molar mass of = 40.3 g/mol
Putting values in equation 1, we get:
[tex]\text{Mass of }MgO=(0.011mol\times 40.3g/mol)=0.443g[/tex]
The percent yield of a reaction is calculated by using an equation:
[tex]\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100[/tex] ......(2)
Given values:
Actual value of the product = 0.437 g
Theoretical value of the product = 0.443 g
Plugging values in equation 1:
[tex]\% \text{yield}=\frac{0.437g}{0.443g}\times 100\\\\\% \text{yield}=98.64\%[/tex]
Hence, the % yield of the product is 98.64 %
Average of a measurement is calculated by given formula:
[tex]Average=\frac{M_1+M_2}{2}[/tex]
where,
[tex]M_1[/tex] = percentage yield for Trial 1 = 98.54 %
[tex]M_2[/tex] = percentage yield for Trial 2 = 98.64 %
Putting values in above equation, we get:
[tex]Average=\frac{98.54+98.64}{2}=98.59\%[/tex]
Hence, the average percent yield of MgO is 98.59 %.