5.33 text{ g} of CrCl_3 cdot 6H_2O, which is | vedclass

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(D) $1$. Calculate the molar mass of $CrCl_3 \cdot 6H_2O$: $52 + (35.5 	imes 3) + (6 	imes 18) = 52 + 106.5 + 108 = 266.5 	ext{ g/mol}$.$2$. Calcul

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$33$

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(D) $1$. Calculate the molar mass of $CrCl_3 \cdot 6H_2O$: $52 + (35.5 	imes 3) + (6 	imes 18) = 52 + 106.5 + 108 = 266.5 	ext{ g/mol}$.$2$. Calculate the moles of the complex: $	ext{Moles} = \frac{5.33 	ext{ g}}{266.5 	ext{ g/mol}} = 0.02 	ext{ mol}$.$3$. Calculate the molar mass of $AgCl$: $108 + 35.5 = 143.5 	ext{ g/mol}$.$4$. Calculate the moles of $AgCl$ formed: $	ext{Moles} = \frac{8.61 	ext{ g}}{143.5 	ext{ g/mol}} = 0.06 	ext{ mol}$.$5$. The ratio of moles of complex reacted to moles of $AgCl$ formed is $\frac{0.02}{0.06} = \frac{1}{3} \approx 0.3333$.$6$. Expressing this as $	imes 10^{-2}$, we get $33.33 	imes 10^{-2}$. The nearest integer is $33$.

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