Question #118697
A teaspoon of milk of magnesia contains 12.0 mg of magnesium hydroxide. What volume of 0.010 mol/L HCl in a person's stomach would be neutralized by this teaspoon of antacid?
1
Expert's answer
2020-05-28T10:33:06-0400

The reaction of neutralization of magnesium hydroxide by hydrochloric acid is:

Mg(OH)2 + 2HCl\rightarrow MgCl2 + 2H2O.

Therefore, 1 mol of magnesium hydroxide reacts with 2 mol of hydrogen chloride:

n(Mg(OH)2)=n(HCl)2n(Mg(OH)_2) = \frac{n(HCl)}{2} .

The number of the moles of magnesium hydroxide can be calculated as its mass divided by its molar mass (58.32 g/mol):

n(Mg(OH)2)=mM=12.0103 g58.32 g/mol=2.058104 moln(Mg(OH)_2) = \frac{m}{M} = \frac{12.0·10^{-3} \text{ g}}{58.32 \text{ g/mol}} = 2.058·10^{-4}\text{ mol}.

The number of the moles of HCl is then the double of this value:

n(HCl)=n(Mg(OH)2)2=4.115104 moln(HCl) = n(Mg(OH)_2) ·2 = 4.115·10^{-4}\text{ mol} .

Finally, the volume of the HCl solution would be neutralized is:

V=nc=4.115104 mol0.010 mol/L=41.2V = \frac{n}{c} = \frac{4.115·10^{-4}\text{ mol}}{0.010\text{ mol/L}} = 41.2 mL.

Answer: the volume of 0.010 mol/L HCl would be neutralized by a teaspoon of antacid is 41.2 mL.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS