What is the approximate mass of the precipita | vedclass

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(B) Step $1$: Calculate the mass of solutes. Assuming density is $1 	ext{ g/mL}$, mass of $50 	ext{ mL}$ solution is $50 	ext{ g}$.Mass of $AgNO_3

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$7$

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(B) Step $1$: Calculate the mass of solutes. Assuming density is $1 	ext{ g/mL}$, mass of $50 	ext{ mL}$ solution is $50 	ext{ g}$.Mass of $AgNO_3 = 16.9\% 	ext{ of } 50 	ext{ g} = 0.169 	imes 50 = 8.45 	ext{ g}$.Mass of $KCl = 7.45\% 	ext{ of } 50 	ext{ g} = 0.0745 	imes 50 = 3.725 	ext{ g}$.Step $2$: Calculate moles of reactants.Moles of $AgNO_3 = \frac{8.45 	ext{ g}}{169 	ext{ g/mol}} = 0.05 	ext{ mol}$.Moles of $KCl = \frac{3.725 	ext{ g}}{74.5 	ext{ g/mol}} = 0.05 	ext{ mol}$.Step $3$: Reaction equation: $AgNO_3 + KCl ightarrow AgCl(s) + KNO_3$.Since stoichiometric ratio is $1:1$ and moles are equal, both are limiting reagents.Moles of $AgCl$ formed $= 0.05 	ext{ mol}$.Step $4$: Mass of $AgCl = 0.05 	ext{ mol} 	imes 143.3 	ext{ g/mol} = 7.165 	ext{ g} \approx 7 	ext{ g}$.

What is the approximate mass of the precipitate formed when $50 \text{ mL}$ of $16.9\%$ solution of $AgNO_3$ is mixed with $50 \text{ mL}$ of $7.45\%$ $KCl$ solution (in $\text{ g}$)? (Molar mass of $AgNO_3 = 169 \text{ g/mol}$, $KCl = 74.5 \text{ g/mol}$, $AgCl = 143.3 \text{ g/mol}$)

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