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Stoichiometric Calculations

A chloride mi...

Question

A chloride mixture is prepared by grinding together pure $B a C l_{2} .2 H_{2} O$ , $K C l$ and $N a C l$ . The smallest and largest volume of $0.15 M$ $A g N O_{3}$ solution that may be used for complete precipitation of chloride from a $0.3$ g sample of the mixture which may contain any one or all of the constituents are respectively:

A

16.38

mL,

34.18

mL

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B

8.19

mL,

34.18

mL

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C

16.38

L,

34.18

L

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D

16.38

mL,

17.09

L

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Solution

The correct option is A $16.38$ mL, $34.18$ mL
$n_{f}$ for $N a C l = 1$
$n_{f}$ for $K C l = 1$
$n_{f}$ for $A g N O_{3} = 1$
$n_{f}$ for $B a C l_{2} .2 H_{2} O = 2$
Maximum amount of $A g N O_{3}$ will be used when $0.3 g$ mixture contains only $N a C l$ whereas minimum amount is used when $0.3 g$ mixture contains only $B a C l_{2} .2 H_{2} O$

According to law of chemical equivalence:
Equivalents of $N a C l$ consumed = Equivalents of $A g N O_{3}$ consumed.
$1 \times \frac{0.3}{58.5} = 1 \times 0.15 \times V_{A g N O_{3}}$
$V_{A g N O_{3}} = 34.18 m l$

Equivalents of $B a C l_{2} .2 H_{2} O$ consumed = Equivalents of $A g N O_{3}$ consumed.
$2 \times \frac{0.3}{244} = 1 \times 0.15 \times V_{A g N O_{3}}$
$V_{A g N O_{3}} = 16.38 m l$

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