Question

The first step in the reaction of Alka–Seltzer with stomach acid consists of one mole of sodium bicarbonate (NaHCO3) reacting with one mole of hydrochloric acid (HCl) to produce one mole of carbonic acid (H2CO3), and one mole of sodium chloride (NaCl). Using this chemical stoichiometry, determine the number of moles of carbonic acid that can be produced from 5 mol NaHCO3 and 9 mol HCl.

Answer 1

Answer: 5 moles of  $H_2CO_3$ can be produced from 5 mol NaHCO3 and 9 mol HCl.

Explanation:

The balanced chemical reaction is:

$NaHCO_3+HCl\rightarrow H_2CO_3+NaCl$

According to stoichiometry :

1 mole of $NaHCO_3$ use 1 mole of $HCl$

Thus 5 moles of $NaHCO_3$ use=$\frac{1}{1}\times 5=5moles$  of $HCl$

Thus $NaHCO_3$ is the limiting reagent as it limits the formation of product and $HCl$ is the excess reagent.

As 1 mole of $NaHCO_3$ give = 1 mole of $H_2CO_3$

Thus 5 moles of $NaHCO_3$ give =$\frac{1}{1}\times 5=5moles$  of $H_2CO_3$

5 moles of  $H_2CO_3$ can be produced from 5 mol $NaHCO_3$ and 9 mol HCl.