nNa2CO3 = 0,05 mol
Na2CO3 + 2HCl \(\rightarrow\) 2NaCl + CO2 + H2O
\(\Rightarrow\) mdd HCl = \(\dfrac{0,1.36,5.100}{1,825}\) = 200 (g)
\(\Rightarrow\) mdd sau pư = 5,3 + 200 - ( 0,05.44 ) = 203,1 (g)
\(\Rightarrow\) C%dd sau pư = \(\dfrac{0,1.59,5.100}{203,1}\) \(\approx\) 2,92%