A mixture of 10cm3 of CO, 60cm3 of H2 and 25cc of CH4 are mixed with 750cm3 of air [containing 20% oxygen] and ignited. Calculate the composition of the resultant mixture on cooling to room temperature.
Answers
Answer:
Burning something is combining a fuel with oxygen. If you burn a hydrocarbon, you get water (hydrogen and oxygen) and carbon dioxide (carbon and oxygen). If you burn hydrogen gas (H2) then you get water (H2O).
(1) CH4 + 2O2 −→heat→heat CO2 + 2H2O
(2)2H2 + O2 →→ 2H2O
Now we have carbon dioxide and water as the products.
Number of moles of reactants:
(1) 10cm3= 10g of CO2
No. of moles = 1044=.22 moles1044=.22 moles
(2) 60cm3 = 60 g of H2
No . of moles = 602=30 moles602=30 moles
(3)25cm3 = 25g of CH4
No. of moles = 2516=1.5 moles2516=1.5 moles
(4) 20% of 75cm3 of oxygen:
.2 ××750 = 150 cm3 = 150g of O2
No. of moles of O2 = 15032=4.68 moles15032=4.68 moles
For the first reaction,0.22 moles of CH4 and 0.44(2 ××0.22) moles of O2 give 0.22 moles of CO2 and 0.44moles (2 ××0.22) of H2O.
For second reaction,H2 is in excess and now O2 is the limiting reagent,therefore,after reaction 1,O2 left = 4.6-.44=4.16 moles.
therefore,4.16 moles of O2 gives 8.32 (2 ×× 4.16) moles of water. Thus moles of H2 used = 8.32 moles.
Moles of H2 (left) = 30-8.32=21.68 moles.
Moles of CO2 = 0.22(produced) + 0.22 (already present)=0.44 moles
Moles of H2O = 0.44(from reaction 1) + 8.32 = 8.76 moles.
Therefore, 21.68:0.44:8.76 of composition of resultant mixture.
MARK AS BRAINLIEST