Question

Assuming an efficiency of 49.30%, calculate the actual yield of magnesium nitrate formed from 133.9 g of magnesium and excess copper(II) nitrate.

Mg + Cu(NO3)2 ---> Mg(NO3)2 + Cu

____ g

Answer 1

Molar mass of Mg = 24.31 g/mol

mass(Mg)= 133.9 g

number of mol of Mg,

n = mass of Mg/molar mass of Mg

=(133.9 g)/(24.31 g/mol)

= 5.508 mol

Balanced chemical equation is:

Mg + Cu(NO3)2 ---> Mg(NO3)2 + Cu

Molar mass of Mg(NO3)2,

MM = 1*MM(Mg) + 2*MM(N) + 6*MM(O)

= 1*24.31 + 2*14.01 + 6*16.0

= 148.33 g/mol

According to balanced equation

mol of Mg(NO3)2 formed = (1/1)* moles of Mg

= (1/1)*5.508021

= 5.508021 mol

mass of Mg(NO3)2 = number of mol * molar mass

= 5.508*1.483*10^2

= 817 g

% yield = actual mass*100/theoretical mass

49.3= actual mass*100/817

actual mass=403 g

Answer: 403 g