Answer :
Answer: The theoretical yield and percent yield of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] is 3.93 g and 30.53 % respectively
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of [tex]CoCl_2.6H_2O[/tex] = 4.00 g
Molar mass of [tex]CoCl_2.6H_2O[/tex] = 238 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }CoCl_2.6H_2O=\frac{4.00g}{238g/mol}=0.0168mol[/tex]
The chemical equation for the reaction of [tex]CoCl_2.6H_2O[/tex] to form [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] follows:
[tex]CoCl_2.6H_2O+4NH_3\rightarrow [Co(NH_3)_4(H_2O)_2]Cl_2+4H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of [tex]CoCl_2.6H_2O[/tex] produces 1 mole of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex]
So, 0.0168 moles of [tex]CoCl_2.6H_2O[/tex] will produce = [tex]\frac{1}{1}\times 0.0168=0.0168mol[/tex] of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex]
Now, calculating the mass of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] from equation 1, we get:
Molar mass of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] = 234 g/mol
Moles of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] = 0.0168 moles
Putting values in equation 1, we get:
[tex]0.0168mol=\frac{\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2}{234g/mol}\\\\\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2=(0.0168mol\times 234g/mol)=3.93g[/tex]
To calculate the percentage yield of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex], we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] = 1.20 g
Theoretical yield of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] = 3.93 g
Putting values in above equation, we get:
[tex]\%\text{ yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=\frac{1.20g}{3.93g}\times 100\\\\\% \text{yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=30.53\%[/tex]
Hence, the theoretical yield and percent yield of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] is 3.93 g and 30.53 % respectively