Question

What volume of 0.600 M HCl is required to react completely with 2.50 g of sodium hydrogen carbonate?

\({\rm{NaHC}}{{\rm{O}}_{3(aq)}}{\rm{ + HC}}{{\rm{l}}_{(aq)}}{\rm{ }} \to {\rm{ NaC}}{{\rm{l}}_{(aq)}}{\rm{ + C}}{{\rm{O}}_{2(aq)}}{\rm{ + }}{{\rm{H}}_2}{{\rm{O}}_{({\rm{l}})}}\)

Short answer

50 ml of hydrochloric acid will be required to react with 0.2.50 g sodium hydrogen carbonate.

Step-by-step solution

Number of moles of NaHCO3

The number of moles is the ratio of the given mass and molar mass of the compound. As the given mass of \({\rm{NaHC}}{{\rm{O}}_3}\)is 2.50 g and molecular mass of \({\rm{NaHC}}{{\rm{O}}_3}\)is \({\rm{23 + 1 + 12 + 16 }} \times {\rm{ 3}} = {\rm{ 84 g/mol}}\)

Therefore, the number of moles = \(\frac{{2.50}}{{84}}{\rm{ = 0}}{\rm{.03 moles}}\)

Now, as 1 moles of hydrochloric acid is required to react with one mole of \({\rm{NaHC}}{{\rm{O}}_3}\), thus the number of moles of nitric acid will be \({\rm{0}}{\rm{.03 }} \times {\rm{ 1 = 0}}{\rm{.03 moles}}\)

Volume of hydrochloric acid

The volume of hydrochloric acid, according to the mole concept can be calculated as:

\({\rm{Concentration = }}\frac{{{\rm{Moles}}}}{{{\rm{Volume}}}}\)

It is given that the concentration of hydrochloric acid is 0.600 M and the number of moles are 0.03 moles, therefore volume of hydrochloric acid will be

\({\rm{Volume = }}\frac{{0.03}}{{0.600}}{\rm{ = 50 mL}}\)