White phosphorus (P4) burns in excess oxygen to give P4O10, which is then treated with water: P4O10 + 6H2O -> 4H3PO4. Starting from 12.4 g of phosphorus and only 5.4 g of water (P = 31, H = 1, O = 16), the maximum mass of H3PO4 formed is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3mc048

White phosphorus ( $P X_{4}$ ) burns in excess oxygen to give $P X_{4} O X_{10}$ , which is then treated with water: $P X_{4} O X_{10} + 6 H X_{2} O 4 H X_{3} PO X_{4}$ . Starting from $12.4 g$ of phosphorus and only $5.4 g$ of water (P = 31, H = 1, O = 16), the maximum mass of $H X_{3} PO X_{4}$ formed is:

n (P) = 12.4/31 = 0.40

, so

n (P X_{4} O X_{10}) = 0.10

mol, which needs

0.6

mol water. Water available

= 5.4/18 = 0.30

mol, so water is limiting.

H X_{3} PO X_{4}

= 0.30 \times 4/6 = 0.20

mol, mass

= 0.20 \times 98 = 19.6 g