From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be___×10^–1 g.

The correct answer is 95.

To find the maximum amount of acetanilide that can be prepared from aniline, we can write down the chemical equation for the acetylation of aniline:
$C_6{H}_{5}N{H}_2+(C{H}_{3}CO{)}_{2}O\to C_6{H}_{5}NHCOC{H}_3+C{H}_{3}COOH$
From the balanced chemical equation, we can see that 1 mole of aniline produces 1 mole of acetanilide.

Step 1: Calculate the molar masses of the reactants and products.
Molar mass of aniline ($C_6{H}_{7}N$):
$\text{Molar mass}=(6\times 12)+(7\times 1)+(14\times 1)=72+7+14=93\text{g/mol}$

Molar mass of acetanilide ($C_8{H}_{9}NO$):
$\text{Molar mass}=(8\times 12)+(9\times 1)+(14\times 1)+(16\times 1)=96+9+14+16=135\text{g/mol}$

Step 2: Calculate the moles of aniline in 6.55 g.
$\text{Moles of aniline}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{6.55}{93}\approx 0.07043\text{mol}$

Step 3: Calculate the maximum mass of acetanilide formed.
Since 1 mole of aniline yields 1 mole of acetanilide, the moles of acetanilide produced will also be approximately 0.07043 mol.
$\text{Mass of acetanilide}=\text{Moles}\times \text{Molar mass}=0.07043\times 135\approx 9.508\text{g}$

Step 4: Convert the mass into the required format.
We need the answer in the form of $x\times {10}^{-1}\text{g}$:
$9.508\text{g}=95.08\times {10}^{-1}\text{g}$
Rounding to the nearest integer gives 95.