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Calculate the number of grams of reactant left over when 28.35 grams of silver reacts with 50.0 liters of air at STP which is 20.9476 % oxygen by volume. The density of air at STP is 1.292 g/L.
attemt at a solution:
m = dv = 64.6 g of air
moles silver =28.35/108=0.26
moles of air : n= PV/RT = (1atm)(50L)/(.0821)(298K)=2mol
moles of air left= 2 - 0.26 =1.74 mol

iam stuck on this need some help please

  • chemistry -

    Its the oxygen reacting with silver, not air; therefore, I think I would get the oxygen out of the air first. If the air is that percent oxygen, I think I would multiply 50.0 L air x 0.209476 = ?? L oxygen. Convert L oxygen to mols oxygen and go from there.
    Ag + O2 ==> Ag2O
    You have mols Ag. You have mols oxygen. Determine the limiting reagent. That reagent will be used completely. The other will have some remaining. You can determine the amount used by the amount of Ag2O formed. Then subtract to see how much remains.

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