$M g (O H)_{2}$ in the form of milk of magnesia is used to neutralize excess stomach acid. How many moles of stomach acid can be neutralized by 1 g of $M g (O H)_{2}$ ?
(Molar mass of $M g (O H)_{2} = 58.33$ )

0.0171

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0.0343

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0.686

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1.25

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Solution

The correct option is A 0.0343
A balanced equation is:
$M g (O H)_{2} + 2 H C l \to M g C l_{2} + 2 H_{2} O$
Number of moles of stomach acid neutralized
$= 2 \times$ number of moles of $M g (O H)_{2}$
$= 2 \times \frac{1}{58.33} = 0.0343$

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