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## Q. 3.11

Consider again the reaction discussed in Example 3.10:

2Sb(s) + 3$I_{2}$(s) →a 2Sb$I_{3}$(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.

## 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!