Chapter 9, Problem 44A

Interpretation Introduction

Interpretation:

The student's percent yield of MgO needs to be deduced given that the theoretical and experimental yield were 1.44 g and 1.23 g respectively.

Concept Introduction:

• 'Yield refers to the amount of product obtained from a reaction.
• 'Theoretical yield' refers to the amount of product obtained from the molar ratio in the balanced equation whereas the 'actual yield' represents the product recovered from an experimental procedure.
• 'Percent yield' is a ratio of the actual and theoretical yields.

  $\text{\% Yield = }\frac{\text{Actual yield (or experimental yield)}}{\text{Theoretical yield}}\text{×100 ------(1)}$

Answer to Problem 44A

% yield of MgO = 85.4%

Explanation of Solution

Given:

Theoretical yield of MgO = 1.44 g

Actual yield of MgO = 1.23 g

Calculations:

Based on equation (1):

  $\begin{array}{l}\text{\% Yield = }\frac{\text{Actual yield (or experimental yield)}}{\text{Theoretical yield}}\text{×100}\\ \text{=}\frac{1.23}{1.44}\times 100=85.4\%\end{array}$

Conclusion

Therefore, the yield of MgO is 85.4%