_ $F e (O H)_{3} +$ _ $(O H)^{-} +$ _ $C l O^{-} \to$ _ ${F e O_{4}}^{2 -} +$ _ $C l^{-} +$ _ $H_{2} O$
When the equation above is balanced, how many grams of $H_{2} O$ are produced when $1 m o l e$ of $F e (O H)_{3}$ is consumed?

18 g

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27 g

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36 g

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45 g

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Solution

The correct option is C $45 g$
$2 F e (O H)_{3} + 4 (O H)^{-} + 3 C l O^{-} \to {2 F e O_{4}}^{2 -} + 3 C l^{-} + 5 H_{2} O$ is the balanced reaction.
$∴$ mass of $H_{2} O$ produced per mole of $F e (O H)_{3} = 18 \times 5 / 2 = 45 g$

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