Answer: The percent yield of zinc carbonate is 5.91 %
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} ZnCO_3=\frac{12.6 g}{125.4g/mol}=0.100moles[/tex]
[tex]ZnI_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Na_2CO_3[/tex] is the excess reagent.
[tex]Na_2CO_3+ZnI_2\rightarrow 2NaI+ZnCO_3[/tex]
According to stoichiometry :
1 mole of produce = 1 mole of
Thus 1.7 moles of [tex]ZnI_2[/tex] will produce=[tex]\frac{1}{1}\times 1.7=1.7moles[/tex] of [tex]ZnCO_3[/tex]
Theoretical yield of [tex]ZnCO_3=moles\times {\text {Molar mass}}=1.7moles\times 125.4g/mol=213.2g[/tex]
percentage yield = [tex]\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100=\frac{12.6g}{213.2g}\times 100=5.91\%[/tex]
