Question

Number of moles of NH3 produced if 140 gm of N2 reacts with 40 gm of hydrogen.
(Given % yield of reaction is 50%)
(A) 12
(B) 10
(C) 5
(D) 6​

Answer 1

Answer: 6 mole

Explanation:

  N₂    +   3H₂    ⟶ 2NH₃

140g        40g  

​moles of N₂ = 140/28 = 5 mol

moles of H₂ = 40/2 = 20 mol

n of H₂    =   20 mol   =    6.66   (hence LR is H₂)

s. coff.             3

⇒ moles of NH₃ = 6.66 × 2 = 13.32

⇒ %yield is given 50%

hence; actual moles of (50 × 13.32 ) / 100 = 6.66

Answer 2

Answer:

5

Explanation:

N2 + 3H2 ----> 2NH3

Given,

mass of N2 = 140g

and mass of H2 = 40g

NOW.

no. of mole of N2 = give mass/ molar mass

= 140/28 = 5 mole

and

no. of mole of H2 = 40/2 = 20 mole

Now. finding Limiting Reagant

for N2 = no. of mole / stoichiometry coefficient

N2 = 5/1 = 5

and

H2 = 20/3 = 6.6

Here Limiting Reagant is N2

Now compare N2 and NH3

N2 + 3H2 = 2NH3

1mol of N2 produce 2 mol of NH3

5 mol of N2 produce 2×5 = 10 mol of NH3

A/Q

given % yield of reaction is 50%

as yield of Reaction is 50% so 5 mole of NH3 will produce

50/100×10 = 5 mole

5 mole of NH3 will produce....