Question

Na2CO3(aq) + CuCl2(aq) =CuCO3(s) + 2NaCl(s) what is the theoretical yield of cuco3

Answer 1

First of all, the equation is incorrect. NaCl will not be formed in a solid form as well as CuCO₃ would, most probably, not be formed:

$2Na_2CO_3(aq)+2CuCl_2(aq)+H_2O \rightarrow (CuOH)_2CO_3(s)+CO_2(g)+4NaCl(aq)$

Now, let assume that the reaction still follows the initial equation. Thus, the theoretical yield is the relation between the amount in moles of the formed CuCO₃ precipitate and the starting reagent consumed in the reaction (Na₂CO₃ or CuCl₂):

$\gamma(CuCO_3)=\frac{n(CuCO_3)}{n(Na_2CO_3)}=\frac{n(CuCO_3)}{n(CuCl_2)}$

$\gamma(CuCO_3)=\frac{m(CuCO_3)M(Na_2CO_3)}{m(Na_2CO_3)M(CuCO_3)}=\frac{m(CuCO_3)M(CuCl_2)}{m(CuCl_2)M(CuCO_3)}$