Question

An old antacid commercial claimed that each tablet of their product could neutralize 47 times its mass in stomach acid. The active ingredient in the antacid tablet, NaAl ( OH ) 2 CO 3 , NaAl(OH)2CO3, reacts with the HCl HCl in stomach acid according to the balanced reaction here. NaAl ( OH ) 2 CO 3 + 4 HCl ⟶ NaCl + AlCl 3 + 3 H 2 O + CO 2 NaAl(OH)2CO3+4HCl⟶NaCl+AlCl3+3H2O+CO2 How many moles of HCl HCl can a 1.24 1.24 g antacid tablet neutralize if the tablet contains 0.296 0.296 g of the active ingredient?

Answer 1

Answer : The number of moles of HCl neutralize is, 0.00824 moles.

Explanation :

The given balanced chemical reaction is:

$NaAl(OH)_2CO_3+4HCl\rightarrow NaCl+AlCl_3+3H_2O+CO_2$

First we have to calculate the moles of $NaAl(OH)_2CO_3$

$\text{Moles of }NaAl(OH)_2CO_3=\frac{\text{Mass of }NaAl(OH)_2CO_3}{\text{Molar mass of }NaAl(OH)_2CO_3}$

Molar mass of $NaAl(OH)_2CO_3$ = 144 g/mol

$\text{Moles of }NaAl(OH)_2CO_3=\frac{0.296g}{144g/mol}=0.00206mol$

Now we have to calculate the moles of HCl.

From the balanced chemical reaction we conclude that,

As, 1 mole of $NaAl(OH)_2CO_3$ react with 4 moles of HCl

So, 0.00206 mole of $NaAl(OH)_2CO_3$ react with $0.00206\times 4=0.00824$ moles of HCl

Thus, the number of moles of HCl neutralize is, 0.00824 moles.