Answered

Consider the following intermediate reactions.
CH4(g)+2O2->CO2(g)+2H2O(g) ΔH1=-802 kJ
2H2O(g)->2H2O(I) ΔH2=-88 kJ
The overall chemical reaction is as follows.
CH4(g)+2O2(g)->CO2(g)+2H2O(I) ΔH2=-890 kJ
What is the correct enthalpy diagram using the Hess law for this system?

Consider the following intermediate reactions. CH4(g)+2O2->CO2(g)+2H2O(g) ΔH1=-802 kJ 2H2O(g)->2H2O(I) ΔH2=-88 kJ The overall chemical reaction is as follows. C class=
Consider the following intermediate reactions. CH4(g)+2O2->CO2(g)+2H2O(g) ΔH1=-802 kJ 2H2O(g)->2H2O(I) ΔH2=-88 kJ The overall chemical reaction is as follows. C class=
Consider the following intermediate reactions. CH4(g)+2O2->CO2(g)+2H2O(g) ΔH1=-802 kJ 2H2O(g)->2H2O(I) ΔH2=-88 kJ The overall chemical reaction is as follows. C class=
Consider the following intermediate reactions. CH4(g)+2O2->CO2(g)+2H2O(g) ΔH1=-802 kJ 2H2O(g)->2H2O(I) ΔH2=-88 kJ The overall chemical reaction is as follows. C class=

Answer :

2.1648 kg of CH4 will generate 119341 KJ of energy.

Explanation:

Write down the values given in the question

CH4(g) +2 O2 → CO2(g) +2 H20 (g)

ΔH1 = - 802 kJ

2 H2O(g)→2 H2O(I)

ΔH2= -88 kJ

The overall chemical reaction is

CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ

CH4 +2 O2 → CO2 +2 H20

(1mol)+(2mol)→(1mol+2mol)

Methane (CH4) = 16 gm/mol

oxygen (O2) =32 gm/mol

Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O

which generate 882 KJ /mol

Therefore to produce 119341 KJ of energy

119341/882 = 135.3 mol

to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require

=135.3 *16

=2164.8 gm

=2.1648 kg of CH4

2.1648 kg of CH4 will generate 119341 KJ of energy

trainofthought

Answer:

The answer is A on edge

Explanation:

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