Question

the mass of N2F4 produce by the reaction of 1 mole of NH3 and 2.5mole F2 is 35.6 g. hence, the percentage yield of given reaction is
2NH3 + 5F2 = N2F4 + 6HF ​

Answer 1

Answer:

2NH3 + 5F2 = N2F4 + 6HF

i.e., 2g NH3 = 0.12 moles NH3

which needs 0.29 moles F2, i.e., 11.15g F2.

Then N2F4 will be 0.587 moles = 6.107g

but

here we have F2 = 8g only,

so the product will be 4.38g N2F4

Therefore, % yield = (4.38/6.107)x100 = 71.72

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