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Question

The number of moles of NaCl required to react with an excess of AgNO3 to produce 14.34 g of AgCl is:

(Molar mass of AgCl is 143.43 g mol−1)

A
0.1 moles
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B
0.2 moles
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C
0.05 moles
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D
0.025 moles
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Solution

The correct option is A 0.1 moles
AgNO3 + NaCl AgCl + NaNO3

Molar Mass of AgCl = 143.32 g

Number of moles of AgCl in 14.34 g of AgCl = Given mass of AgClMolar Mass of AgCl
=14.34143.32=0.1 mol

1 mole of NaCl produces 1 mole of AgCl

So, the number of moles of NaCl that produces 0.1 mol of AgCl
= 0.1 mol

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