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The complete reaction of 6.8 g of ${ FeCl }_{ 2 }$ (MM = 126.75 g/mol) produced 1.34 g of ${ CrCl }_{ 3 }$ (MM = 158.35 g/mol).

Calculate the percent yield for the reaction.

$${ Na }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+ 6{ FeCl }_{ 2 }+14HCl \rightarrow { 2CrCl }_{ 3 }+{ 6FeCl }_{ 3 }+{ 7H }_{ 2 }O+2NaCl$$

A

19.7 %

B

73.9 %

C

47.3%

D

15.8%

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