From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be ___ × \( 10^{-1} \) g.
Correct Answer: 95
Solution and Explanation
Reaction:
\[\text{Aniline (C}_6\text{H}_5\text{NH}_2) + \text{CH}_3\text{COCl} \rightarrow \text{Acetanilide (C}_6\text{H}_5\text{NHCOCH}_3\text{)}\]
From stoichiometry, 93 g of aniline gives 135 g of acetanilide.
So, for 6.55 g of aniline:
\[\text{Amount of acetanilide} = \frac{135}{93} \times 6.55 = 9.5 \, \text{g}\]
In scientific notation: \( 9.5 = 95 \times 10^{-1} \, \text{g}. \)
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