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Industrial Preparation of Ammonia

15. THe mass ...

Question

15. THe mass of N2F4 produced by 2.0g NH3 and 8.0g F2 is 3.56g according to the reaction 2NH3 + 5F2 = N2F4 +6HF THEN FIND PERCENT PER YIELD

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2 N H_{3} (g) + 5 F_{2} \to N_{2} F_{4} + 6 H F

3.56 g N_{2} F_{4}

N H_{3}

F_{2}

N_{2} F_{4}

2 N H_{3} + 5 F_{2} \to N_{2} F_{4} + 6 H F

In the reaction,

2 $N H_{3}$ + $F_{2}$ $⟶$ $N_{2} F_{4}$ +6HF

16.85g of $N_{2} F_{4}$ is obtained by mixing 2g $N H_{3}$ and 8g $F_{2}$ . The percentage yield of the production is:

(N_{2} H_{4})

C l N H_{2} + 2 N H_{3} \to N_{2} H_{4} + N H_{4} C l

C l N H_{2}

N H_{3}

N_{2} H_{4}

(N_{2} H_{4})

C l N H_{2} + 2 N H_{3} \to N_{2} H_{4} + N H_{4} C l

C l N H_{2}

N H_{3}

N_{2} H_{4}

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