Question

Use the References to access important values if needed for this question. A 14.1 g sample of an aqueous solution of hydrochloric acid contains an unknown amount of the acid. IF 22.5 mL of 1.22 M sodium hydroxide are required to neutralize the hydrochloric acid, what is the percent by mass of hydrochloric acid in the mixtual by mass Submit Answer Retry Entire Group 9 more group attempts remaining

Answer 1

Balanced chemical equation is:

NaOH + HCl ---> NaCl + H2O

lets calculate the mol of NaOH

volume , V = 22.5 mL

= 2.25*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 1.22*2.25*10^-2

= 2.745*10^-2 mol

According to balanced equation

mol of HCl reacted = (1/1)* moles of NaOH

= (1/1)*2.745*10^-2

= 2.745*10^-2 mol

This is number of moles of HCl

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

use:

mass of HCl,

m = number of mol * molar mass

= 2.745*10^-2 mol * 36.46 g/mol

= 1.001 g

Use:

Mass % of HCl = mass of HCl * 100 / mass of solution

= 1.001 * 100 / 14.1

= 7.10 %

Answer: 7.10 %