From $6.55 \mathrm{~g}$ of aniline, the maximum amount of acetanilide that can be prepared will be ________ $\times 10^{-1} \mathrm{~g}$.

<p>To determine the maximum amount of acetanilide that can be prepared from 6.55 g of aniline, we need to use stoichiometry. Let's go through the process step by step.</p> <p>1. <strong>Molecular weights calculation:</strong></p> <p>The molecular weight of aniline (C₆H₅NH₂) is calculated as follows:</p> <p>$$\text{Molecular weight of aniline} = 6 \times 12 + 5 \times 1 + 14 + 2 \times 1 = 93 \text{ g/mol}$$</p> <p>The molecular weight of acetanilide (C₈H₉NO) is calculated as follows:</p> <p>$$\text{Molecular weight of acetanilide} = 8 \times 12 + 9 \times 1 + 14 + 16 = 135 \text{ g/mol}$$</p> <p>2. <strong>Mole calculation:</strong> <p>Moles of aniline:</p> <p>$$\text{Moles of aniline} = \frac{\text{Mass of aniline}}{\text{Molecular weight of aniline}} = \frac{6.55 \text{ g}}{93 \text{ g/mol}} = 0.0704 \text{ mol}$$</p> <p>3. <strong>Stoichiometry of the reaction:</strong> <p>The reaction between aniline and acetic anhydride to form acetanilide follows a 1:1 mole ratio.</p> <p>4. <strong>Mass calculation:</strong> <p>Theoretical mass of acetanilide formed:</p> <p>$$\text{Mass of acetanilide} = \text{Moles of aniline} \times \text{Molecular weight of acetanilide} = 0.0704 \text{ mol} \times 135 \text{ g/mol} = 9.504 \text{ g}$$</p> <p>5. <strong>Convert to the desired unit:</strong> <p>Given the unit required is $\times 10^{-1} \mathrm{~g}$, we express 9.504 g as:</p> <p>$9.504 \text{ g} = 95.04 \times 10^{-1} \text{ g}$</p> <p>Therefore, the maximum amount of acetanilide that can be prepared from 6.55 g of aniline is $95.04 \times 10^{-1} \mathrm{~g}$.</p>