Assuming an efficiency of 26.20%, calculate the actual yield of magnesium nitrate formed from 128.6 g of magnesium and excess copper(II) nitrate.
Mg + Cu (NO3)2 --------> Mg (NO3)2 + Cu
_____ grams
Molar mass of Mg = 24.31 g/mol
mass of Mg = 128.6 g
we have below equation to be used:
number of mol of Mg,
n = mass of Mg/molar mass of Mg
=(128.6 g)/(24.31 g/mol)
= 5.29 mol
we have the Balanced chemical equation as:
Mg + Cu(NO3)2 ---> Mg(NO3)2 + Cu
Molar mass of Mg(NO3)2 = 1*MM(Mg) + 2*MM(N) + 6*MM(O)
= 1*24.31 + 2*14.01 + 6*16.0
= 148.33 g/mol
From balanced chemical reaction, we see that
when 1 mol of Mg reacts, 1 mol of Mg(NO3)2 is formed
mol of Mg(NO3)2 formed = (1/1)* moles of Mg
= (1/1)*5.29
= 5.29 mol
we have below equation to be used:
mass of Mg(NO3)2 = number of mol * molar mass
= 5.29*1.483*10^2
= 7.847*10^2 g
% yield = actual mass*100/theoretical mass
26.2= actual mass*100/784.6663
actual mass=205.6 g
Answer: 205.6 g
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