Single Choice

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?

A29.16 g
Correct Answer
B34.3 g
C58.33 g
D68.66 g

Solution

Neutralization reaction is:
$$Mg(OH)_2 + 2HCl \rightarrow 2H_2O + MgCl_2$$
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 $$Mg(OH)_2=24.3+(16+1)\times 2=58.3\ g$$
0.5 mol of $$Mg(OH)_2 =0.5 \times 58.3=29.15\ g$$
option A is correct

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