An orgo lab student named Bob wanted to synthesize an isoamyl acetate ester because it smells like bananas and he really likes to sniff it. He heated 7.50 grams of isoamyl alcohol (3-methyl-1-butanol) with 22.5 grams of acetic acid (ethanoic acid) and 4.50 grams of sulfuric acidto prepare his favorite ester. Based on these numbers, how many grams of isoamyl acetateshould he theoretically obtain

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Answer:

The answer is "11.07 g".

Explanation:

Isoamyl alcohol is a reagent restriction

Isoamyl alcohol Moles: 

[tex]= \frac{7.5}{88.15} \\\\ =0.085[/tex]

Moles only with the shape of isoamyl acetate are equivalent to numbers.

Isoamyl acetate grams:

[tex]= 0.085 \times 130.19\\\\ = 11.07 \ g[/tex]