How many grams of $M g (O H)_{2}$ will be needed to neutralize 25 ml of stomach acid if stomach acid is 0.10 M HCl?

0.0171

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0.0343

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0.0729

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1.25

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Solution

The correct option is A 0.0729
Moles of $H C l = 0.10 \times 0.025 L = 0.0025$
$2 H C l + M g (O H)_{2} \to M g C l_{2} + 2 H_{2} O$
Moles of $M g (O H)_{2} = \frac{0.0025}{2} = 0.00125$
Mass of $M g (O H)_{2} = 0.00125 m o l \times 58.327 g / m o l = 0.0729 g$

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