Question #24259
How many moles of Ag can be formed when 5.0 mol of Zn react with 4.2 mol of AgNO3?
Expert's answer
Firstly, let's find the amount of substance for each product:
n(AgNO3)= 4.2 mol
n(Zn) = 5 mol
Next reaction:
2AgNO3 + Zn = Zn(NO3)2 + 2Ag
zinc is in excess, so we'll use silver nitrate for calculations:
n(Ag) = n(AgNO3)
m(Ag) = n(Ag)*Ar(Ag)
m(Ag) = 4.2 mol * 108 g/mol
m(Ag) = 453,6 g
n(AgNO3)= 4.2 mol
n(Zn) = 5 mol
Next reaction:
2AgNO3 + Zn = Zn(NO3)2 + 2Ag
zinc is in excess, so we'll use silver nitrate for calculations:
n(Ag) = n(AgNO3)
m(Ag) = n(Ag)*Ar(Ag)
m(Ag) = 4.2 mol * 108 g/mol
m(Ag) = 453,6 g
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