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  1. Problems
  2. /Coke burns in a limited supply of air so that all of the oxygen forms carbon monoxide: 2C + O₂ →…
MediumChemistryHigh School (11–12)The p-Block Elements › Groups 13 and 14: Boron and Carbon FamiliesNumericalA-LevelIBAEEE (Amrita)AMUEEEAP EAPCETBCECE
+15 more exams
COMEDK UGETGUJCETJCECEJKCETKCETKEAMKIITEELPUNESTMET (Manipal)MHT-CETOJEESRMJEEETS EAPCETUPESEATVITEEE

Coke burns in a limited supply of air so that all of the oxygen forms carbon monoxide: 2C + O₂ →…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Coke burns in a limited supply of air so that all of the oxygen forms carbon monoxide: 2C+O2→2CO2\text{C} + \text{O}_{2} \to 2\text{CO}2C+O2​→2CO. Air is 21.0%21.0\%21.0% oxygen and 79.0%79.0\%79.0% nitrogen by volume, and nitrogen takes no part. Find the percentages by volume of carbon monoxide and nitrogen in the gas that leaves. Then find the heat released when the carbon monoxide in 1.00 m31.00\ \text{m}^31.00 m3 of the gas burns to carbon dioxide, with ΔH=−283.0 kJ/mol\Delta H = -283.0\ \text{kJ/mol}ΔH=−283.0 kJ/mol for CO, at 273 K273\ \text{K}273 K and 1 atm1\ \text{atm}1 atm. Use 22.4 L/mol22.4\ \text{L/mol}22.4 L/mol and 1 m3=1000 L1\ \text{m}^3 = 1000\ \text{L}1 m3=1000 L.

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. The gas from 100 volumes of air. Oxygen: 21.021.021.0 volumes, which gives 2×21.0=42.02 \times 21.0 = 42.02×21.0=42.0 volumes of CO. Nitrogen: 79.079.079.0 volumes.

Step 2. Percentages. The total is 42.0+79.0=121.042.0 + 79.0 = 121.042.0+79.0=121.0 volumes. CO: 42.0/121.0=34.7%42.0 / 121.0 = 34.7\%42.0/121.0=34.7%. Nitrogen: 79.0/121.0=65.3%79.0 / 121.0 = 65.3\%79.0/121.0=65.3%.

Step 3. Moles of CO in 1.00 m31.00\ \text{m}^31.00 m3. 0.347×1000 L/22.4 L/mol=15.5 mol0.347 \times 1000\ \text{L} / 22.4\ \text{L/mol} = 15.5\ \text{mol}0.347×1000 L/22.4 L/mol=15.5 mol.

Step 4. Heat. 15.5×283.0=4385 kJ=4.39 MJ15.5 \times 283.0 = 4385\ \text{kJ} = 4.39\ \text{MJ}15.5×283.0=4385 kJ=4.39 MJ.

Answer: 34.7%34.7\%34.7% CO, 65.3%65.3\%65.3% N2_22​ and 4.39 MJ4.39\ \text{MJ}4.39 MJ.

Check. The gas has twice the volume of CO as of O2 used, so it is larger than the air that went in, by a factor of 1.211.211.21.

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