What should be the weight and moles of AgCl p | vedclass

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(A) The balanced chemical equation is: $AgNO_3 + HCl 	o AgCl + HNO_3$Calculate the moles of reactants:Moles of $AgNO_3 = \frac{30 \ g}{170 \ g/mol} \

Vedclass · Accepted Answer

$14.35$

Vedclass · Answer

(A) The balanced chemical equation is: $AgNO_3 + HCl 	o AgCl + HNO_3$Calculate the moles of reactants:Moles of $AgNO_3 = \frac{30 \ g}{170 \ g/mol} \approx 0.176 \ mol$Moles of $HCl = 	ext{Molarity} 	imes 	ext{Volume (L)} = 0.20 \ M 	imes 0.500 \ L = 0.1 \ mol$Since $HCl$ has fewer moles,it is the limiting reagent.According to the stoichiometry,$1 \ mol$ of $HCl$ produces $1 \ mol$ of $AgCl$.Therefore,$0.1 \ mol$ of $AgCl$ is produced.Weight of $AgCl = 	ext{moles} 	imes 	ext{molar mass} = 0.1 \ mol 	imes 143.5 \ g/mol = 14.35 \ g$.

What should be the weight and moles of $AgCl$ precipitate obtained on adding $500 \ mL$ of $0.20 \ M$ $HCl$ to $30 \ g$ of $AgNO_3$ solution (in $g$)? (Molar mass of $AgNO_3 = 170 \ g/mol$)

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