What is the weight of the precipitate formed | vedclass

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(A) The chemical reaction is: $AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$.First,calculate the moles of reactants:Molar mass of $AgNO_3 =

Vedclass · Accepted Answer

$7.17$

Vedclass · Answer

(A) The chemical reaction is: $AgNO_3(aq) + NaCl(aq) ightarrow AgCl(s) + NaNO_3(aq)$.First,calculate the moles of reactants:Molar mass of $AgNO_3 = 108 + 14 + 3 	imes 16 = 170 \ g/mol$.Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$.Mass of $AgNO_3$ in $50 \ mL$ of $16.9 \%$ solution = $(16.9 / 100) 	imes 50 = 8.45 \ g$.Moles of $AgNO_3 = 8.45 / 170 = 0.0497 \approx 0.05 \ mol$.Mass of $NaCl$ in $50 \ mL$ of $5.8 \%$ solution = $(5.8 / 100) 	imes 50 = 2.9 \ g$.Moles of $NaCl = 2.9 / 58.5 = 0.0495 \approx 0.05 \ mol$.Since the stoichiometry is $1:1$,both reactants are consumed completely.Moles of $AgCl$ formed = $0.05 \ mol$.Molar mass of $AgCl = 108 + 35.5 = 143.5 \ g/mol$.Weight of $AgCl$ precipitate = $0.05 \ mol 	imes 143.5 \ g/mol = 7.175 \ g \approx 7.17 \ g$.

What is the weight of the precipitate formed when $50 \ mL$ of $16.9 \%$ $AgNO_3$ solution is mixed with $50 \ mL$ of $5.8 \%$ $NaCl$ solution (in $g$)?

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