1.
nFe2O3 = 0,1 mol
Fe2O3 + 6HCl \(\rightarrow\) 2FeCl3 + 3H2O
0,1.........0,6............0,3
\(\Rightarrow\) mdd HCl = \(\dfrac{0,6.36,5.100}{10}\) = 219 (g)
\(\Rightarrow\) C%dd sau phản ứng = \(\dfrac{0,2.162,5.100}{16+219}\) \(\approx\) 13,8%
2.
mNaOH = 20 (g)
\(\Rightarrow\) nNaOH = 0,5 mol
NaOH + HCl \(\rightarrow\) NaCl + H2O
\(\Rightarrow\) C%dd HCl = \(\dfrac{0,5.36,5.100}{245}\) \(\approx\) 7,4%
\(\Rightarrow\) C%dd sau phản ứng = \(\dfrac{0,5.58,5.100}{\left(0,5.40\right)+245}\) \(\approx\) 11,03%