Question

How many grams of phosphoric acid $\left(H_{3}P{O}_4\right)$ would be required to neutralize $58g$ of magnesium hydroxide?

• A. $65.3g$
• B. $70.2g$
• C. $85.0g$
• D. $112.0g$

Answer 1

Answer: (A)

$65.3g$

The ratio of the weight of phosphoric acid to its equivalent weight is equal to the ratio of the weight of magnesium hydroxide to its equivalent weight.

${\left(\frac{W}{E}\right)}_{H_{3}P{O}_4}={\left(\frac{W}{E}\right)}_{Mg(OH{)}_2}$

Substituting values in the above expression, we get-

$\frac{W}{98}\times 3=\frac{58}{58}\times 2$

$W=65.3g$

Hence, $65.33$ grams of phosphoric acid would be required to neutralize $58$ g of magnesium hydroxide.

Hence, option A is correct.