I have an assigenment where i need to calculate the lover heating value of  the combustion of 100 mol of a gas mixture (MJ/Nm³)

The gas mixture is made of in mol%:
91.1% CH₄
4.7% C₂H₆
1.7% C₃H₈
1.4% C₄H₁₀
0.6% N₂
0.5% CO₂

the enthalpy of the combustion reaction of methane, ethane, propane and butane, are given:
Methane combustion : -804,2 kJ/mol
Ethane combustion : -1428,9 kJ/mol
Propane combustion : -2045,5 kJ/mol
butane combustion : -2663.6 kJ/mol

we are also told the gasmixture behaves like an idealgas.

The solution in the book is given to be 38,8 MJ/m³ but i can't seem to get the right answer no matter what i do basketball stars

my main theory to calculating it was to calculate the ΔH of the combustion reaction using the enthalpy of the 4 gas combustions given by adding them together  and multiplying with their respective mol% and then dividing it with the normal volume of an idealgas 22,42 m³/kmol which gives me a solution of 38,93 MJ/M³ which is close to the solution but not quite it.

Does anyone know how what i am missing?

38 / 5.000
This case is really rare 

Under standard conditions, the more likely product of the reaction between iron (Fe) and copper(II) chloride (CuCl₂) is FeCl₂ (iron(II) chloride), with iron in the +2 oxidation state:

Fe + CuCl₂ → FeCl₂ + Cu
However, under specific conditions (e.g., high temperature or a strong oxidizing environment), it’s possible to form FeCl₃ (iron(III) chloride) with iron in the +3 oxidation state:

Fe + CuCl₂ → FeCl₃ + Cu
So, the actual product depends on the reaction conditions space waves 

This week at the lab I tried to synthesize [Co(Me)(py)(DMG)2] this is a summary of the procedure:

Day 1:
 
  - Dissolved 2.7g of DMG in 100mL of EtOH with heat and stirring. 
  - Added 2.5g of CoCl₂·6H₂O, forming a dark green solution. 
  - Added 2mL of py and let it cool to room temperature. 
  - Passed air through the mixture for 30 minutes, turning the solution light brown (three phases were observed: a dark brown layer on top, a light brown suspension in the middle, and a small dark green phase at the bottom). 
  - Filtered under vacuum and washed with water, then ethanol, and finally ether, obtaining a brown powder. 

---

- **Day 2** 

  - Assembled a system with a Liebig condenser and a 2-neck round bottom flask, purging it with nitrogen for 5 minutes. 
  - Added 10mL of MeOH and deoxygenated it for another 5 minutes. 
  - Added 0.8g of the compound obtained on Day 1 (presumably [Co(Cl)(py)(DMG)₂]) and 2mL of methyl iodide, forming a dark-colored solution. 
  - Slowly added 0.4g of NaBH₄. Upon addition, an orange layer formed on top but disappeared once the NaBH₄ was consumed. 
  - Added a small quantity of water to precipitate the product. 
  - Filtered the solution, obtaining a brown-orange powder. This brown was different from the one obtained on Day 1 but still brown (maybe incomplete or partial reaction?) 

  - Repeated the procedure of day 2 step by step, starting with 0.8g of [Co(Cl)(py)(DMG)₂] provided by another lab mate (which appeared identical to the compound I obtained on day 1). This time, the expected orange powder, likely [Co(Me)(py)(DMG)₂], was obtained

I guess I did something wrong on day 1. The only different step on day 2 from the first try to the second try was that on the second the stirring was more intense (not a lot tho).

Any ideas? If cobalt had not been oxidized on day 1 it should've remained green instead of brown right?

Sounds like a simple stoichiometry - first step is to find out reagents and products, then you need to balance the reaction equation.

For balancing same compound can't be present on both sides of the equation.

Why iodine from potassium iodate has diffrent colour than from iodide?

Iodate solution looks like mud, brown/dark beige.
Iodide solution looks goldish - yellow/brown, even strong violet and the begining.

Why is that?

Iodate solution made of: starch + acid + iodide
Iodide solution made of: starch + acid + peroxide.