530 g of $Na X_{2} CO X_{3}$ is fully converted to $NaOH$ by boiling with excess slaked lime, and all the $NaOH$ is then treated with excess chlorine in hot concentrated solution. The mass of $NaClO X_{3}$ formed is (Na 23, C 12, O 16, Cl 35.5; assume complete reactions)

Answer: A

Na X_{2} CO X_{3} + Ca (OH) X_{2} 2 NaOH + CaCO X_{3}

530/106 = 5.00

mol carbonate gives 10.00 mol NaOH. Hot concentrated alkali:

3 Cl X_{2} + 6 NaOH 5 NaCl + NaClO X_{3} + 3 H X_{2} O

, so 10.00/6 = 1.667 mol

NaClO X_{3}

(106.5 g/mol), mass 177.5 g. The value 372.5 g would correspond to

NaOCl

from the cold dilute reaction.