Question

When a mixture of 10 moles of SO2 and 15 moles of O2 was passed over catalyst, 8 moles of So
was formed. How many moles of SO2 and O2did not enter into combination

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Answer 1

Answer:

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.......