Chapter 3
Q. 3.11
Consider again the reaction discussed in Example 3.10:
2Sb(s) + 3I_{2}(s) →a 2SbI_{3}(s)
Suppose that in part (a) the percent yield is 78.2%. How many grams of SbI3 are formed?
ANALAYSIS | |
From Example 3.10a, theoretical yield (1.20 mol) percent yield (78.2%) |
Information given: |
mass SbI_{3} actually obtained | Asked for: |
STRATEGY
1. Substitute into Equation 3.3.
percent yield = \frac{\text{experimental yield}}{\text{theoretical yield}} × 100% (3.3)
% yield = \frac{\text{actual yield}}{\text{theoretical yield}} × 100%
2. Your answer will be the actual yield in moles. Convert to grams.
Step-by-Step
Verified Solution
78.2% = \frac{actual yield}{1.20 mol} × 100%; actual yield = 0.938 mol SbI_{3} | actual yield |
0.938 mol SbI_{3} × \frac{502.5 g SbI_{3}}{1 mol SbI_{3}} = 472 g | mass SbI_{3} |
END POINT
If your actual yield is larger than your theoretical yield, something’s wrong!