In basic medium $Cr (OH) X_{3}$ is oxidised by chlorate: $Cr (OH) X_{3} + ClO X_{3} X^{-} + OH X^{-} CrO X_{4} X^{2 -} + Cl X^{-} + H X_{2} O$ . In the balanced equation (smallest whole numbers) find the sum of the coefficients of $OH X^{-}$ and $H X_{2} O$ . Give the answer as an integer.

Numerical answer type, tolerance 0.1.

Answer: 9

Cr loses 3 electrons (+3 to +6) and Cl gains 6 (+5 to -1), so

Cr (OH) X_{3}

ClO X_{3} X^{-}

= 2 : 1. Charge balance needs 4

OH X^{-}

on the left, and O/H balance gives 5

H X_{2} O

2 Cr (OH) X_{3} + ClO X_{3} X^{-} + 4 OH X^{-} 2 CrO X_{4} X^{2 -} + Cl X^{-} + 5 H X_{2} O

. Sum of coefficients of

OH X^{-}

and

H X_{2} O

= 4 + 5 = 9.