White phosphorus is burned: $P X_{4} + 5 O X_{2} P X_{4} O X_{10}$ , using $12.4 g$ of $P X_{4}$ ( $124 g/mol$ ) and $12.8 g$ of $O X_{2}$ . The $P X_{4} O X_{10}$ formed (the first step runs to completion, consuming the limiting reagent fully) is then hydrated, $P X_{4} O X_{10} + 6 H X_{2} O 4 H X_{3} PO X_{4}$ , with $90%$ yield. The mass of $H X_{3} PO X_{4}$ ( $98 g/mol$ ) obtained, in grams to two decimals, is

Numerical answer type, answer in g, tolerance 0.02.

Answer: 28.22 g

P X_{4}

= 0.1 mol needs 0.5 mol

O X_{2}

; only 0.4 mol is present, so

O X_{2}

limits.

P X_{4} O X_{10}

= 0.4/5 = 0.08 mol.

H X_{3} PO X_{4}

= 4 x 0.08 x 0.90 = 0.288 mol x 98 = 28.224 g.

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