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Byju's Answer

Standard XI

Statistics

Pie Diagram

In the reacti...

Question

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:

81.28%

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71.2%

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68%

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77%

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Solution

The correct option is D
77%

$F_{2}$ is limiting reagent.

38g of flourine produces 104g of $N_{2} H_{4}$

8g of flourine produces $\frac{8 \times 104}{38}$ =21.89g of $N_{2} H_{4}$

%yield = $\frac{16.85}{21.89}$ $\times$ 100 =77%

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In the reaction, 2NH3+F2⟶N2F4+6HF 16.85g of N2F4 is obtained by mixing 2gNH3 and 8g F2. The percentage yield of the production is:

The correct option is D 77% F2 is limiting reagent. 38g of flourine produces 104g of N2H4 8g of flourine produces 8×10438=21.89g of N2H4 %yield = 16.8521.89×100 =77%

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:

The correct option is D
77%

$F_{2}$ is limiting reagent.

38g of flourine produces 104g of $N_{2} H_{4}$

8g of flourine produces $\frac{8 \times 104}{38}$ =21.89g of $N_{2} H_{4}$

%yield = $\frac{16.85}{21.89}$ $\times$ 100 =77%