Statement 1: $1.435$ g of $A g C l$ will be precipitated when $200$ ml of $5$ N HCl is mixed to a solution containing $1.7$ g of $A g N O_{3} (A g C l = 143.5)$ .
Statement 2: The limiting reagent is $A g N O_{3}$ .

Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation of Statement 1.

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Statement 1 is True, Statement 2 is True, Statement 2 is not a correct explanation of Statement 1.

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Statement 1 is true, Statement 2 is False.

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Statement 1 is False, Statement 2 is True.

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Solution

The correct option is A Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation of Statement 1.
The reaction is given as $A g N O_{3} + H C l \to A g C l + H N O_{3}$
For one mole of precipitaion of $A g C l$ we need one mole of $A g N O_{3}$ and one mole of $H C l$
For the given problem, moles of $H C l = N V = 0.2 \times 5 = 1 m o l e$
And moles of $A g N O_{3}$ is $\frac{1.7}{170} = 0.01$ .
So, $A g N O_{3}$ is the limiting reagent and $A g C l$ is equal to $0.01$ moles = $1.435$ g

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Statement 1: 1.435 g of AgCl will be precipitated when 200 ml of 5 N HCl is mixed to a solution containing 1.7 g of AgNO3(AgCl=143.5).Statement 2: The limiting reagent is AgNO3.

Statement 1: $1.435$ g of $A g C l$ will be precipitated when $200$ ml of $5$ N HCl is mixed to a solution containing $1.7$ g of $A g N O_{3} (A g C l = 143.5)$ .
Statement 2: The limiting reagent is $A g N O_{3}$ .