Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 4 (Q.No. 20)
20.
A gaseous mixture contains 14 kg of N2, 16 kg of O2 and 17 kg of NH3. The mole fraction of oxygen is
Discussion:
8 comments Page 1 of 1.
Rishi said:
1 decade ago
Mole fraction of O2=16/14+16+17=0. 33.
(2)
Gabriel Rosado said:
1 decade ago
I think something is wrong.
Mass fraction of O2= 16/(14+16+17) = 0.34.
Mole fraction of O2= (16/32)/(14/28+16/32+17/17) = 0.25.
Right?
Mass fraction of O2= 16/(14+16+17) = 0.34.
Mole fraction of O2= (16/32)/(14/28+16/32+17/17) = 0.25.
Right?
(1)
Deven sonj said:
9 years ago
The correct answer should be 0.25.
(1)
Hanadi said:
8 years ago
0.25 is correct answer.
(1)
Deepak said:
4 years ago
.25 is the correct answer.
(1)
Ankur varma said:
10 years ago
Yes the correct answer is 0.25 only, which is not given in the option only.
@Rishi they have asked for mole fraction of oxygen and not the "MASS FRACTION". If they could have ask for mass fraction they 0.34 is the correct answer.
@Rishi they have asked for mole fraction of oxygen and not the "MASS FRACTION". If they could have ask for mass fraction they 0.34 is the correct answer.
Hatdu said:
4 years ago
Explain the answer.
TANMAY SHAH said:
2 years ago
Mass [gm] = M.W. × Moles.
By putting the given values in above equation we get the number of moles of,
N2 = 500 moles,
O2 = 500 moles,
NH3 = 1000 moles,
So, we get the mole fraction of O2 is 0.25.
By putting the given values in above equation we get the number of moles of,
N2 = 500 moles,
O2 = 500 moles,
NH3 = 1000 moles,
So, we get the mole fraction of O2 is 0.25.
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