Respuesta :
Answer:
%yield of [tex]PbSO_{4}[/tex] is 33%
Explanation:
20.0 mL of 0.80 M [tex]Pb(NO_{3})_{2}[/tex] = [tex]\frac{20.0\times 0.80}{1000}[/tex] moles of
= 0.016 moles of [tex]Pb(NO_{3})_{2}[/tex]
= 0.016 moles of [tex]Pb^{2+}[/tex]
Balanced reaction: [tex]Pb^{2+}+SO_{4}^{2-}\rightarrow PbSO_{4}[/tex]
According to balanced equation, 1 mol of [tex]Pb^{2+}[/tex] produces 1 mol of [tex]PbSO_{4}[/tex]
So, 0.016 mol of [tex]Pb^{2+}[/tex] produces 0.016 mol of [tex]PbSO_{4}[/tex]
As sodium sulphate remains in excess amount therefore lead nitrate is the limiting reagent.
So, theoretical yield of [tex]PbSO_{4}[/tex] = 0.016 mol
Molar mass of [tex]PbSO_{4}[/tex] = 303.26 g/mol
So, theoretical yield of [tex]PbSO_{4}[/tex] = [tex](0.016\times 303.26)g=4.85g[/tex]
%yield of [tex]PbSO_{4}[/tex] = [(actual yield)/(theoretical yield)][tex]\times 100[/tex]%
=( [tex]\frac{1.6}{4.85}\times 100[/tex]) %
= 33%
