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Q. $HCl$ is produced in the stomach which can be neutralised by $Mg(OH)_{2}$ in the form of milk of magnesia. How much $Mg(OH)_{2}$ is required to neutralise one mole of stomach acid?

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Solution:

$Mg \left(OH\right)_{2} + 2HCl \to MgCl_{2} +2H_{2}O$
No. of moles of $Mg(OH)_{2}$ required for $2$ moles of $HCl = 1 $
No. of moles of $Mg(OH)_{2}$ required for $1$ mole of $HCl = 0.5 $
Mass of $0.5\, mol$ of $Mg(OH)_{2} = 58.33 \times 0.5 = 29.16\, g$