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A student mixes 2.83 mL of benzoyl chloride with excess 15 M NH4OH to produce 1.95 g of benzamide. What is the percent yield of this student's experiment

User XWang
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Answer:

Step-by-step explanation:

The reaction of benzoyl chloride with NH₄OH to produce benzamide is:

Benzoyl chloride + ammonia → Benzamide + NH₄Cl

Molar mass of benzoyl chloride: 140.57 g/mol. Density 1.21g/mL

Molar mass benzamide: 121.14g/mol.

To know percent yield you must know the theoretical yield of the reaction (How many grams are produced assuming a yield of 100%). Percent yield will be (Actual yield / Theoretical Yield) ₓ 100

Moles of 2.83mL of benzoyl chloride are:

2.83mL ₓ (1.21g/mL) ₓ (1mol / 140.57g) = 0.02436 moles of benzoyl chloride.

As 1 mole of benzoyl chloride produce 1 mole of benzamide (Theoretical yield), theoretical moles of benzamide produced are 0.02436. In mass:

0.02436 moles ₓ (121.14g / mol) = 2.95g of benzoyl chloride

As there are produced just 1.95, percent yield is:

(1.95g / 2.95g) ₓ 100 = 66.1%

User Jared Egan
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