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Standard XII

Chemistry

Percentage Composition

For the react...

Question

For the reaction given below:
$P b (N O_{3})_{2} (a q) + 2 N a C l (a q) \to$ $P b C l_{2} + N a N O_{3} (a q)$
The volume of $0.75 M$ $P b (N O_{3})_{2}$ in mL required to react completely with $1.0$ L of $0.225 M$ $N a C l$ solution is :

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Solution

$1.0$ L of $0.225$ M $N a C l$ corresponds to $1.00 \times 0.225$ moles.
Let $V$ be the volume of lead(II) chloride.
The number of moles of $P b C l_{2} = V (r e q u i r e d) \times 0.750$ .
$2$ moles of $N a C l$ will react with $1$ mole of lead nitrate.
Hence, $0.225$ moles of $N a C l$ will react with $0.1125$ moles of lead nitrate.
$V \times 0.750 = 0.1125$
Thus, the volume is $0.15$ L or $150$ mL.

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For the reaction given below:Pb(NO3)2(aq)+2NaCl(aq) → PbCl2+NaNO3(aq)The volume of 0.75M Pb(NO3)2 in mL required to react completely with 1.0 L of 0.225M NaCl solution is :

For the reaction given below:
$P b (N O_{3})_{2} (a q) + 2 N a C l (a q) \to$ $P b C l_{2} + N a N O_{3} (a q)$
The volume of $0.75 M$ $P b (N O_{3})_{2}$ in mL required to react completely with $1.0$ L of $0.225 M$ $N a C l$ solution is :