Chapter 11.4, Problem 35SSC

Interpretation Introduction

Interpretation − $$ Percent yield of the reaction of magnesium with excess oxygen needs to be determined.

Concept introduction − Percent yield is the ration of actual yield to theoretical yield.

Answer to Problem 35SSC

Percent yield is 87.9%

Explanation of Solution

Mass of Mg

= Mass of Mg and crucible − Mass of empty crucible

= $38.06-35.67$

= $3.48g$

Step 1: Balance the chemical equation

$2{\text{Mg}}_{\text{(s)}}{\text{ + O}}_{2(\text{g)}}\to {\text{ 2MgO}}_{(\text{s)}}$

Make mass to mole conversion

For conversion from mass to mole, we divide mass by molecular mass, so in this case moles of Mg are:

$2.39\text{g Mg}\times \frac{\text{1 mole Mg}}{24.31\text{g Mg}}=0.0983\text{ mole Mg}$

Step 3: Make moles to mole conversion

As we know from the balanced chemical equation that 2 moles Mg produces 2 moles MgO. Therefore, moles of MgO produced by 0.0983 moles of Mg:

$0.0983\text{ mole Mg}\times \frac{2\text{ mole of MgO}}{2\text{ mole of Mg}}=0.0983\text{ mole Mg}$

Step 4: Makes mole to mass conversion

For conversion of moles to mass, we multiply mole by molecular mass, so in this case mass of Mg is:

$0.0983\text{ mole Mg}\times 40.31g\text{ of Mg = 3}\text{.96g of Mg}$

$\begin{array}{l}\text{ Percent yield = }\frac{\text{Actual yield}}{\text{Theoratical yield}}\times 100\\ \text{ = }\frac{\text{3}\text{.48 g MgO}}{\text{3}\text{.96 g MgO}}\times 100\\ \text{ = 87}\text{.9\%}\end{array}$

Conclusion

Thus, percent yield is 87.9%