In the reaction,
$2 N H_{3} (g) + 5 F_{2} \to N_{2} F_{4} + 6 H F$
$3.56 g N_{2} F_{4}$ is obtained by mixing 2 g $N H_{3}$ and 8 g $F_{2}$ . The percentage yield of the production is:

81.28%

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71.2%

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68%

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79%

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Solution

The correct option is A 81.28%
$2 N H_{3} (g) 34 g + 5 F_{2} 190 g \to N_{2} F_{4} 104 g + 6 H F$

Amount of $N_{2} F_{4}$ formed by $2 g N H_{3} = \frac{104}{34} \times 2 = 6.12$

Amount of $N_{2} F_{4}$ formed by $8 g F_{2} = \frac{104}{190} \times 8 = 4.38$

$N_{2} F_{4}$ will be limiting and actual amount of product is 3.56 g

$∴$ % yield $= \frac{A c t u a l a m o u n t o f p r o d u c t}{C a l c u l a t e d a m o u n t o f p r o d u c t} \times 100$

$= \frac{3.56}{4.38} \times 100 = 81.28$ %

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Similar questions

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:

For the reaction
$A l + O_{2} \to A l_{2} O_{3}$
$54 g$ of $A l$ is reacted with $67.4$ litre of $O_{2}$ at STP. $76.5 g$ of $A l_{2} O_{3}$ was obtained. The percentage yield of the product obtained is

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For the reaction
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(The equation is not balanced)
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In the reaction, 2NH3(g)+5F2→N2F4+6HF3.56gN2F4 is obtained by mixing 2 g NH3 and 8 g F2. The percentage yield of the production is:

The correct option is A 81.28% 2NH3(g)34g+5F2190g→N2F4104g+6HFAmount of N2F4 formed by 2gNH3=10434×2=6.12Amount of N2F4 formed by 8gF2=104190×8=4.38N2F4 will be limiting and actual amount of product is 3.56 g ∴ % yield =ActualamountofproductCalculatedamountofproduct×100=3.564.38×100=81.28%

In the reaction,
$2 N H_{3} (g) + 5 F_{2} \to N_{2} F_{4} + 6 H F$
$3.56 g N_{2} F_{4}$ is obtained by mixing 2 g $N H_{3}$ and 8 g $F_{2}$ . The percentage yield of the production is: