Question

From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be ___ × $10^{-1}$ g.

Answer 1

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: $9.5 = 95 \times 10^{-1} \, \text{g}.$