What weight of silver chloride will be precipitated when a solution containing 4.77 g of $NaCl$ is added to a solution of 5.77 g of $AgNO_{3}?(Na = 23,Cl = 35.5,Ag = 108,N = 14$ and $O = 16)$

Question

Dr. Abubacker · Accepted Answer

(b)
According to the equation,
$NaCl + AgNO_{3} \longrightarrow NaNO_{3} + AgCl$
No. of moles of $NaCl = \frac{4.77}{58.5} = 0.08154$
No. of moles of $AgNO_{3} = \frac{5.77}{170} = 0.03394$
Thus, $AgNO_{3}$ is the limiting reagent in the reaction.
Now, applying $POAC$ for $Ag$ (as $Ag$ atoms are conserved in the reaction)
Moles of $Ag$ in $AgNO_{3} =$ moles of $Ag$ in $AgCl$
Or $1 	imes moles\ of\ AgNO_{3} = 1 	imes moles\ of\ AgCl$
Or $0.03394 	imes 143.4(for\ AgCl) = 4.87g$

Q. What weight of silver chloride will be precipitated when a solution containing 4.77 g of $NaCl$ is added to a solution of 5.77 g of $AgNO_{3}?(Na = 23,Cl = 35.5,Ag = 108,N = 14$ and $O = 16)$

Explanation

Reviewed by Dr. Abubacker

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