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 mL of $H C l_{(a q)}$ with a density of 1.15 g/ml and 30.1% HCl by mass, how many grams of $C l_{2} (g)$ are produced?
$K_{2} C r_{2} O_{7} + H C l \to C r C l_{3} + C l_{2} + H_{2} O + K C l$
(Atomic weights: K = 39, Cr = 52, Cl = 35.5)

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68.16g

45.64g

34.08g

53g

answer is C.

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Detailed Solution

$K_{2} C r_{2} O_{7} + 14 H C l \to 2 C r C l_{3} + 3 C l_{2} + 7 H_{2} O + 2 K C l$

$n C r_{2} O_{7} 2 - = 49 \times 0.96 \times \frac{1}{294} = 0.16 w H C l = 325 \times 1.15 \times 0.301 = 112.5 g n H C l = \frac{112.5}{36.5} = 3.08$ Limiting reagent is dichromate. $w C l_{2} = 31 \times 0.16 \times 71 = 35.01 g$

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