What volume of $0.1 M H C l$ will be oxidized by $1 m L$ of $0.1 M K M n O_{4}$ ?

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Solution

The balanced reaction between $H C l$ and $K M n O_{4}$
$2 K M n O_{4} + 16 H C l ⟶ 2 M n C l_{2} + 5 C l_{2} + 2 K C l + 8 H_{2} O$
Normality of $K M n O_{4} = 0.1 \times 5 = 0.5$
Volume of $K M n O_{4} = 1 m l$
Normality of $H C l = 0.1 \times 1 = 0.1$
Volume of $H C l =$ ?
$\Rightarrow N_{1} V_{1} = N_{2} V_{2}$
$\Rightarrow 0.5 \times 1 = 0.1 \times V_{2}$
$\Rightarrow V_{2} = \frac{0.5}{0.1} = 5 m l$
$∴$ Volume of $H C l$ needed is $5 m l$ .

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What volume of 0.1 M HCl will be oxidized by 1 mL of 0.1 M KMnO4?

The balanced reaction between HCl and KMnO42KMnO4+16HCl⟶2MnCl2+5Cl2+2KCl+8H2ONormality of KMnO4=0.1×5=0.5Volume of KMnO4=1mlNormality of HCl=0.1×1=0.1Volume of HCl= ?⇒N1V1=N2V2⇒0.5×1=0.1×V2⇒V2=0.50.1=5ml∴ Volume of HCl needed is 5ml .

What volume of $0.1 M H C l$ will be oxidized by $1 m L$ of $0.1 M K M n O_{4}$ ?