The mass of AgCl precipitated when a solution | vedclass

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(A) The balanced chemical equation is:$NaCl + AgNO_3 ightarrow AgCl + NaNO_3$Molar mass of $AgNO_3 = 108 + 14 + (3 	imes 16) = 170 \ g/mol$Molar ma

Vedclass · Accepted Answer

$2.87$

Vedclass · Answer

(A) The balanced chemical equation is:$NaCl + AgNO_3 ightarrow AgCl + NaNO_3$Molar mass of $AgNO_3 = 108 + 14 + (3 	imes 16) = 170 \ g/mol$Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$Molar mass of $AgCl = 108 + 35.5 = 143.5 \ g/mol$Given amounts:$NaCl = 11.70 \ g$$AgNO_3 = 3.4 \ g$From the stoichiometry,$1 \ mol$ of $AgNO_3$ reacts with $1 \ mol$ of $NaCl$.$170 \ g$ of $AgNO_3$ requires $58.5 \ g$ of $NaCl$.$3.4 \ g$ of $AgNO_3$ requires $\frac{58.5}{170} 	imes 3.4 = 1.17 \ g$ of $NaCl$.Since we have $11.70 \ g$ of $NaCl$ (which is more than $1.17 \ g$),$AgNO_3$ is the limiting reagent.The amount of $AgCl$ produced depends on the limiting reagent $(AgNO_3)$:$170 \ g$ of $AgNO_3$ produces $143.5 \ g$ of $AgCl$.$3.4 \ g$ of $AgNO_3$ produces $\frac{143.5}{170} 	imes 3.4 = 2.87 \ g$ of $AgCl$.Thus,the mass of $AgCl$ precipitated is $2.87 \ g$.

The mass of $AgCl$ precipitated when a solution containing $11.70 \ g$ of $NaCl$ is added to a solution containing $3.4 \ g$ of $AgNO_3$ is [Atomic mass of $Ag=108$,$Na=23$,$Cl=35.5$,$N=14$,$O=16$] (in $g$)

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