The mass of benzanilide obtained from the benzoylation reaction of 5.8 g of aniline, if yield of product is 82%, is ____ g (nearest integer) .
(Given molar mass in g $$mol^{-1}$$ H : 1, C: 12, N: 14, O : 16)

We are asked to find the mass of benzanilide obtained from the benzoylation of 5.8 g of aniline with 82% yield.

The benzoylation of aniline (Schotten-Baumann reaction) produces benzanilide according to the equation $$\text{C}_6\text{H}_5\text{NH}_2 + \text{C}_6\text{H}_5\text{COCl} \rightarrow \text{C}_6\text{H}_5\text{CONHC}_6\text{H}_5 + \text{HCl}$$ and since one mole of aniline yields one mole of benzanilide, the stoichiometry is 1:1.

The molar mass of aniline ($$\text{C}_6\text{H}_5\text{NH}_2$$) is calculated as $$6(12) + 5(1) + 14 + 2(1) = 72 + 5 + 14 + 2 = 93 \, \text{g/mol}$$ and the molar mass of benzanilide ($$\text{C}_6\text{H}_5\text{CONHC}_6\text{H}_5$$) is $$13(12) + 11(1) + 14 + 16 = 156 + 11 + 14 + 16 = 197 \, \text{g/mol}$$.

The number of moles of aniline is given by $$\text{Moles of aniline} = \frac{5.8}{93} = 0.06237 \, \text{mol}$$.

Since the reaction is 1:1, the theoretical moles of benzanilide are also 0.06237 mol, and the theoretical mass is $$\text{Theoretical mass} = 0.06237 \times 197 = 12.287 \, \text{g}$$.

Applying the 82% yield gives an actual mass of $$\text{Actual mass} = \text{Theoretical mass} \times \frac{\text{yield}}{100} = 12.287 \times \frac{82}{100}$$ and thus $$\text{Actual mass} = 12.287 \times 0.82 = 10.075 \, \text{g}$$.

Finally, rounding to the nearest integer gives $$\approx 10 \, \text{g}$$.

The answer is 10.