Ta có: \(2\mathop {Cl}\limits^{ - 1} \to \mathop {C{l_2}}\limits^0 + 2e\)
A. \(\mathop {Mn}\limits^{ + 4} + 2e \to \mathop {Mn}\limits^{ + 2} \)
\( \Rightarrow {n_{C{l_2}}} = {n_{Mn{O_2}}} = 1mol\)
B. \(\mathop {Mn}\limits^{ + 7} + 5e \to \mathop {Mn}\limits^{ + 2} \)
\( \Rightarrow {n_{C{l_2}}} = \dfrac{5}{2}{n_{Mn{O_2}}} = 2,5mol\)
C. \(\mathop {Cl}\limits^{ + 5} + 6e \to \mathop {Cl}\limits^{ - 1} \)
\( \Rightarrow {n_{C{l_2}}} = 3{n_{KCl{O_3}}} = 3mol\)
D. \(2\mathop {Cl}\limits^0 + 2e \to 2\mathop {Cl}\limits^{ - 1} \)
\( \Rightarrow {n_{C{l_2}}} = {n_{CaOC{l_2}}} = 1mol\)
=> Chọn C
