Question

Answer 1

a) n_{HCl} = \dfrac{10,95}{36,5} = 0,3(mol)$

$2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O$

$n_{Cl_2} = \dfrac{5}{16}n_{HCl} = 0,09375(mol)$

$V_{Cl_2} = 0,09375.22,4 = 2,1(lít)$

b)

$MnO_2 + 4HCl \to MnCl_2 + Cl_2 + H_2O$

$n_{Cl_2} = \dfrac{1}{4}n_{HCl} = 0,075(mol)$
$V_{Cl_2} = 0,075.22,4 = 1,68(lít)$