Answer:
[tex]m_{HCl}=2.19gHCl[/tex]
Explanation:
Hello,
In this case, for the reaction between hydrochloric acid and sodium hydroxide, for 2.4 g of base, we can compute the neutralized grams of acid by applying the 1:1 molar ratio between them and their molar masses, 36.45 g/mol and 40 g/mol respectively as shown below by stoichiometry:
[tex]m_{HCl}=2.4gNaOH*\frac{1molNaOH}{40gNaOH}*\frac{1molHCl}{1molNaOH}*\frac{36.45gHCl}{1molHCl}\\ \\m_{HCl}=2.19gHCl[/tex]
Best regards.