The reaction below can be used as a laboratory method for preparing small quantities of $C l_{2} (g) .$ If a $49 g$ sample that is $96 % K_{2} C r_{2} O_{7}$ by mass is allowed to react with $325 m L$ of $H C l (a q)$ with a density of $1.15 g / m l$ and $30.1 % H C l$ by mass, how many grams of $C l_{2} (g)$ are produced?
$K_{2} C r_{2} O_{7}^{} + H C l \overset{}{\to} C r C l_{3} + C l_{2} + H_{2} O + K C l (n o t b a l a n c e d)$
(Atomic weights: K = 39, Cr = 52, Cl = 35.5)

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$68.16 g$

$45.64 g$

$34.08 g$

$53 g$

answer is C.

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Detailed Solution

$K_{2} C r_{2} O_{7}^{} + 14 H C l \overset{}{\to} 2 C r C l_{3} + 3 C l_{2} + 7 H_{2} O + 2 K C l$
$n_{C r_{2} O_{7}^{2 -}} = 49 \times \frac{96}{100} \times \frac{1}{294} = 0.16$
$w_{H C l} = 325 \times 1.15 \times 0.301 = 112.5 g$
$n_{H C l} = \frac{112.5}{36.5} = 3.08$
Limiting reagent is dichromate.
$w_{C l_{2}} = \frac{3}{1} \times 0.16 \times 71 = 34.08 g$

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