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Question

The addition of 0.15 moles of NaCl to an aqueous solution containing 0.05 moles of AgNO3 results in the formation of an insoluble white precipitate of AgCl. Find the limiting reagent and the amount of AgCl that gets formed.
(Ag=108 u, Cl=35.5 u)

A
NaCl, 7.175 g
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B
AgNO3, 7.175 g
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C
NaCl, 14.350 g
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D
AgNO3, 14.350 g
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Solution

The correct option is B AgNO3, 7.175 g
The corresponding reaction is:
NaCl(aq)+AgNO3(aq)NaNO3(aq)+AgCl(s)
Moles of NaCl=0.15 mol
Moles of AgNO3=0.05 mol
From the balanced stoichiometric equation,
1 mole of NaCl reacts with 1 mole of AgNO3
0.15 moles of NaCl will react with 0.15 moles of AgNO3
So, AgNO3 is the limiting reagent.
As per the reaction, 1 mole of AgNO3 produces 1 mole of AgCl
0.05 moles of AgNO3 will produce 0.05 moles of AgCl
Amount of precipitate formed
=moles of AgCl formed × Molar mass of AgCl
=(0.05×143.5) g
=7.175 g

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