Question

A 523-mL sample of unknown HCl solution reacts completely with Na2CO3 to form 14.1 g CO2. What was the concentration of the HCl solution?

Answer 1

2HCl(aq) + Na2CO3(aq) ---> 2NaCl(aq) + H2O(l) + CO2(g)

n(HCl) = 2 x n(CO2) = (2)(18.1 g / 44.01 g/mol) = 0.823 mol

[HCl] = 0.823 mol / 0.501 L = 1.64 M

Answer 2

bal. rxn.:2 HCl + Na2CO3 -----> CO2 + H2O + NaCl

moles CO2: 11.1 g. CO2/44 g/mole CO2= 0.252
from bal. rxn., 2 moles HCl reacts to form 1 mole of CO2 , therefore 0.504 moles HCl were reacted.
moles= L x M
0.504 moles HCl= 0.613 x M
M HCl= 0.822