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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!