Boric acid (molar mass 62 g/mol) is heated in stages: H3BO3 -> HBO2 -> H2B4O7 -> B2O3 until only B2O3 (molar mass 70 g/mol) remains. The mass of residue from 12.4 g of H3BO3 is

JEE Advanced levelSingle correct optionDifficulty 3 of 3pb181

Boric acid (molar mass 62 g/mol) is heated in stages: $H X_{3} BO X_{3} HBO X_{2} H X_{2} B X_{4} O X_{7} B X_{2} O X_{3}$ until only $B X_{2} O X_{3}$ (molar mass 70 g/mol) remains. The mass of residue from 12.4 g of $H X_{3} BO X_{3}$ is

Moles of

H X_{3} BO X_{3}

= 12.4/62 = 0.20. Two moles of boric acid give one mole of

B X_{2} O X_{3}

(all other boron compounds in the sequence are intermediates), so

B X_{2} O X_{3}

= 0.10 mol, mass = 0.10 x 70 = 7.0 g.