The mass of N2F4 produced by the reaction of 2.0 g of NH3 and 80 g of F2 is 3.56 . what is the percent yield ?2NH3+5F2⟶N2F4+6HF

2NH3+5F2⟶N2F4+6HF2×17g NH3 gives 66 g of N2F42 g will give =6634×2=3.88g N2F4% yield =3.563.88×100=91.75%

The mass of N2F4 produced by the reaction of 2.0 g of NH3 and 80 g of F2 is 3.56 . what is the percent yield ?2NH3+5F2⟶N2F4+6HF

2NH3+5F2⟶N2F4+6HF2×17g NH3 gives 66 g of N2F42 g will give =6634×2=3.88g N2F4% yield =3.563.88×100=91.75%

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The mass of N₂F₄ produced by the reaction of 2.0 g of NH₃ and 80 g of F₂ is 3.56 . what is the percent yield ?
$2 {NH}_{3} + 5 F_{2} ⟶ N_{2} F_{4} + 6 HF$

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Detailed Solution

$\begin{array}{l} 2 {NH}_{3} + 5 F_{2} ⟶ N_{2} F_{4} + 6 HF \\ 2 \times 17 g {NH}_{3} gives 66 g of N_{2} F_{4} \end{array}$
$\begin{array}{l} 2 g will give = \frac{66}{34} \times 2 = 3 .88 g N_{2} F_{4} \\ % yield = \frac{3 .56}{3 .88} \times 100 = 91 .75 % \end{array}$