Consider again the reaction discussed in Example 3.10: 2Sb(s) + 3I2(s) → 2SbI3(s) Suppose that in part (a) the percent yield is 78.2%. How many grams of SbI3 are formed?

Question

Consider again the reaction discussed in Example 3.10:

2Sb(s) + 3[latex]I_{2}[/latex](s) →a 2Sb[latex]I_{3}[/latex](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 [latex]SbI_{3}[/latex] actually obtained	Asked for:

STRATEGY

1. Substitute into Equation 3.3.

percent yield = [latex]\frac{ ext{experimental yield}}{ ext{theoretical yield}}[/latex] × 100% (3.3)

% yield = [latex]\frac{ ext{actual yield}}{ ext{theoretical yield}}[/latex] × 100%

2. Your answer will be the actual yield in moles. Convert to grams.

Accepted Answer

78.2% = [latex]\frac{actual yield}{1.20 mol}[/latex] × 100%; actual yield = 0.938 mol [latex]SbI_{3}[/latex]	actual yield
0.938 mol [latex]SbI_{3}[/latex] × [latex]\frac{502.5 g SbI_{3}}{1 mol SbI_{3}}[/latex] = 472 g	mass [latex]SbI_{3}[/latex]

END POINT

If your actual yield is larger than your theoretical yield, something’s wrong!

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 $SbI_{3}$

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