Let's assume you were given 2.0 g benzil, 2.2 g dibenzyl ketone, 50 mL 95% ethanol and 0.3 g potassium hydroxide to synthesize . You isolated 3.0 g of . What is the % yield

the % yield is 82%

Explanation:

Given the data in the question,

we know that;

Molar mass of benzil is 210.23 g·mol−1

Molar mass of dibenzyl ketone is 210.27 g·mol−1

Molar mass of is 384.5 g·mol−1

Now,

2.0 g benzil = 2 g / 210.23 g·mol−1 = 0.0095 mole

2.2 g dibenzyl ketone = 2.2 / 210.27 = 0.0105 mole

3.0 g of = 3 / 384.5  = 0.0078 mole

Now, the limiting reagent is benzil. 0.0095 mole can reacts wiyh 0.0095 mole of dibenzyl ketone

percentage yield = ( 0.0078 mole / 0.0095 mole ) × 100%

= 0.82 × 100%

= 82%

Therefore, the % yield is 82%

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