Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 2 (Q.No. 1)
1.
One mole of methane undergoes complete combustion in a stoichiometric amount of air. The reaction proceeds as CH4 + 2O2 → CO2 + 2H2O. Both the reactants and products are in gas phase. ΔH°298 = - 730 kJ/mole of methane. Mole fraction of water vapour in the product gases is about
Discussion:
4 comments Page 1 of 1.
Yaregal worku said:
6 years ago
Please explain the solution clearly.
(1)
Bibiu said:
9 years ago
2moles of oxygen = 2 * 79/21 =7 .523 moles of nitrogen (air is used not pure oxygen).
So, composition is:
(a) water -> 2 moles
(b) co2 -> 1 moles
(c) n2 -> 7.523 moles
Total 10.523.
Mole fraction of water 2/10.523 = 0.19.
So, composition is:
(a) water -> 2 moles
(b) co2 -> 1 moles
(c) n2 -> 7.523 moles
Total 10.523.
Mole fraction of water 2/10.523 = 0.19.
(2)
Balveer singh lodha said:
10 years ago
10.524 mole nitrogen.
2.0 mole water.
1.0 mole carbon-dioxide.
A. 0.19.
2.0 mole water.
1.0 mole carbon-dioxide.
A. 0.19.
Shivani said:
10 years ago
Can anyone please explain how to solve this question?
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