Question

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?

• A. 29.16 g
• B. 34.3 g
• C. 58.33 g
• D. 68.66 g

Answer 1

Answer: (A)

29.16 g

Neutralization reaction is:

$Mg(OH{)}_2+2HCl\to 2H_{2}O+MgC{l}_2$

According to the balanced reaction 2 mol of $HCl$ is neutralized by 1 mol of $Mg(OH{)}_2$

$\therefore$ 1 mol of $HCl$ will be neutralised by $\frac{1}{2}$ mol of $Mg(OH{)}_2$

1 mol of $Mg(OH{)}_2$ = Molar mass of $Mg(OH{)}_2=24.3+(16+1)\times 2=58.3g$

0.5 mol of $Mg(OH{)}_2=0.5\times 58.3=29.15g$

option A is correct