If $$279 \text{ g}$$ of aniline is reacted with one equivalent of benzenediazonium chloride, the maximum amount of aniline yellow formed will be ______ g. (nearest integer) (consider complete conversion).

Find the mass of aniline yellow formed from 279 g of aniline with one equivalent of benzenediazonium chloride.

Since aniline reacts with benzenediazonium chloride to give p-aminoazobenzene (aniline yellow) and HCl, the reaction can be written as $$C_6H_5NH_2 + C_6H_5N_2^+Cl^- \rightarrow C_6H_5N=NC_6H_4NH_2 + HCl.$$

The molar mass of aniline ($$C_6H_5NH_2$$) is 93 g/mol, so the number of moles in 279 g is $$\mathrm{Moles} = \frac{279}{93} = 3\ \text{mol}.$$

One equivalent of benzenediazonium chloride reacts with each mole of aniline in a 1:1 ratio, so 3 mol of aniline yields 3 mol of the azo product.

The molar mass of aniline yellow ($$C_{12}H_{11}N_3$$) is $$12\times12 + 11\times1 + 3\times14 = 144 + 11 + 42 = 197\ \text{g/mol},$$ and hence the mass of product is $$3 \times 197 = 591\ \text{g}.$$

The correct answer is 591.