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The mass of $N_{2} F_{4}$ produced by the reaction of $2.0 g$ of $N H_{3}$ and 80 g of $F_{2}$ is $3.56 g$ . what is the percent yield?
$2 N H_{3} + 5 F_{2} \to N_{2} F_{4} + 6 H F$

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$79.0$

$71.2$

$84.6$

91.75

answer is D.

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Detailed Solution

$2 N H_{3} + 5 F_{2} \to N_{2} F_{4} + 6 H F 2 \times 17 g N H_{3} g i v e s 66 g N_{2} F_{4} 2 g w i l l g i v e - \frac{66}{34} \times 2 = 3.88 g N_{2} F_{4} % y i e l d = \frac{3.56}{3.88} \times 100 = 91.75 %$

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