How many grams of phosphoric acid $(H_{2} P O_{4})$ would be needed to neutralise $100$ g of magnesium hydroxide $(M g (O H)_{2})$ ?

66.7

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252

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112.6

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168

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Solution

The correct option is D $112.6$ g
Apply the formula
${(\frac{W}{E q u i v a l e n t w t .})}_{H_{3} P O_{4}} = {(\frac{W}{E q u i v a l e n t w t .})}_{M g {(O H)}_{2}}$
Hence,
$\frac{W}{98 \times 3} = \frac{100}{58 \times 2}$
$∴ W = 112.6 g r a m$

Suggest Corrections

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