From a container having 64g oxygen,11.2L oxygen gas at S.T.P and 6.022×10^23 oxygen atom are removed. Find the mass of the oxygen gas left:
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Hey there!
No. Of moles of O2 present = 64/32=2moles
No.of moles of O2 taken out=11.2/22.4=1/2 mole and 6.022*10^23/6.022*10^23 =1 mole
So total no. Of moles taken out =1 +1/2 moles=3/2 moles
No. Of moles left=2-3/2=1/2mole O2
1 mole O2 conatins 32 g
So 1/2 mole O2 conatins 16 g mass
Hope that helps you :-):-)
No. Of moles of O2 present = 64/32=2moles
No.of moles of O2 taken out=11.2/22.4=1/2 mole and 6.022*10^23/6.022*10^23 =1 mole
So total no. Of moles taken out =1 +1/2 moles=3/2 moles
No. Of moles left=2-3/2=1/2mole O2
1 mole O2 conatins 32 g
So 1/2 mole O2 conatins 16 g mass
Hope that helps you :-):-)
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