From the following reaction sequences, the mass of ${C l}_{2}$ needed to produce $100$ g of $K C l O_{4}$ is (as the nearest integer) :
${C l}_{2} + 2 K O H \to K C l + K C l O + H_{2} O . . . (i)$
$3 K C l O \to 2 K C l + K C l O_{3} . . . (i i)$
$4 K C l O_{3} \to 3 K C l O_{4} + K C l . . . (i i i)$

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Solution

Rewriting the given reaction to have an overall reaction, we get
$4 K C l O_{3} \to 3 K C l O_{4} + K C l$
$12 K C l O \to 8 K C l + 4 K C l O_{3}$
$12 {C l}_{2} + 24 K O H \to 12 K C l + + 12 K C l O + 12 H_{2} O$
Adding all the three, we get
$12 {C l}_{2} + 24 K O H \to 3 K C l O_{4} + 21 K C l + 12 H_{2} O$
$∵ 3 \times 138.5 g K C l O_{4} \equiv 12 \times 71 g {C l}_{2}$
$∴ 100 g K C l O_{4} = \frac{12 \times 71 \times 100}{3 \times 138.5} \equiv 205.05$ g.
So, the nearest integer is $205$ .

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Similar questions

From the following reaction sequence
$C l_{2}$ + 2KOH $\to$ KCl + KClO + $H_{2} O$
3KClO $\to$ 2KCl + $K C l O_{3}$
$4 K C l O_{3}$ $\to$ $3 K C l O_{4} + K C l$

Calculate the mass of chlorine needed to produce 100g of $K C l O_{4}$

From the following reaction sequence
$C l_{2}$ + 2KOH $\to$ KCl + KClO + $H_{2} O$
3KClO $\to$ 2KCl + $K C l O_{3}$
$4 K C l O_{3}$ $\to$ $3 K C l O_{4} + K C l$