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When a mixture of 10 mole of so2 , 15 mole of O2 passed over catalyst 8mole of so3 was formed . How many miles of O2 and so2 did not enter into the reaction. ?

When a mixture of 10 mole of so2 , 15 mole of O2 passed over catalyst 8mole of so3 was formed . How many miles of O2 and so2 did not enter into the reaction. ?

Grade:11

1 Answers

Dev r
39 Points
6 years ago
Molar mass of S=32.06u
Molar mass of O=16.00u
Molar Mass of SO2=64.06u
Mass in grams of 10 moles of SO2=10*32.06=320.6g
Masss in grams of 15 moles O2= 15*32.00=960g
 
 
Mass of S that went in=10*32.06=320.6g
Mass of S that came out in the form of SO3=8*32.06=256.48g
Mass of S that did not take part in the reaction=320.6-256.48=64.12g=2 moles of S
 
Now S took part in the reaction in the form of SO2 so if there were 2 moles of S that didn’t take part in reaction then there were 2 moles of SO2 that did not take part inthe reaction.
 
Now if SO2 is taken out from SO3 Then we have only O so the mass of O2 that was used in the reaction= 8(No. of moles of O in SO3)*16.00(molar mass of oxygen)=128g
 
mass of oxygen that went in=15*32.00(molar mass of O2)=480.00g
Therefore the mass of oxygen that was not used was=480.00-128.00=352g=11 moles of O2
 
Therefore 2 moles of SO2 and 11 moles of O2 did not take part in the reaction

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