Question

The mass of N2F4 produced by the equation of 2.0g of NH3 and 8g of F2 is 3.56g.What is the percent yield?

Answer 1

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

Answer 2

Answer:

81.46%

Explanation:

moles of nh3 = 2/17

moles of nh3 = 2/17moles of F2 = 8/38 = 4/19

moles of nh3 = 2/17moles of F2 = 8/38 = 4/19finding LR:

moles of nh3 = 2/17moles of F2 = 8/38 = 4/19finding LR:(2/17)/2 = 1/17 for Nh3

moles of nh3 = 2/17moles of F2 = 8/38 = 4/19finding LR:(2/17)/2 = 1/17 for Nh3(4/19)/5 = 4/95 for F2

moles of nh3 = 2/17moles of F2 = 8/38 = 4/19finding LR:(2/17)/2 = 1/17 for Nh3(4/19)/5 = 4/95 for F2clearly, f2 is LR

moles of nh3 = 2/17moles of F2 = 8/38 = 4/19finding LR:(2/17)/2 = 1/17 for Nh3(4/19)/5 = 4/95 for F2clearly, f2 is LRhence moles of n2f4 = 4/19 * 1/5 = 4/95

moles of nh3 = 2/17moles of F2 = 8/38 = 4/19finding LR:(2/17)/2 = 1/17 for Nh3(4/19)/5 = 4/95 for F2clearly, f2 is LRhence moles of n2f4 = 4/19 * 1/5 = 4/95 mass of n2f4 produced acc to stoichiometric eq

moles of nh3 = 2/17moles of F2 = 8/38 = 4/19finding LR:(2/17)/2 = 1/17 for Nh3(4/19)/5 = 4/95 for F2clearly, f2 is LRhence moles of n2f4 = 4/19 * 1/5 = 4/95 mass of n2f4 produced acc to stoichiometric eq= 4/95 * 104

moles of nh3 = 2/17moles of F2 = 8/38 = 4/19finding LR:(2/17)/2 = 1/17 for Nh3(4/19)/5 = 4/95 for F2clearly, f2 is LRhence moles of n2f4 = 4/19 * 1/5 = 4/95 mass of n2f4 produced acc to stoichiometric eq= 4/95 * 104 = 4.37 g

moles of nh3 = 2/17moles of F2 = 8/38 = 4/19finding LR:(2/17)/2 = 1/17 for Nh3(4/19)/5 = 4/95 for F2clearly, f2 is LRhence moles of n2f4 = 4/19 * 1/5 = 4/95 mass of n2f4 produced acc to stoichiometric eq= 4/95 * 104 = 4.37 g % y = (3.56/4.37) * 100 ≈ 81.46 %