Question

Theoretically, what mass of [Co(NH3)4(H2O)2]Cl2 could be produced from 4.00 g of CoCl2•6H2O starting material. If 1.20 g of [Co(NH3)4(H2O)2]Cl3 is produced, what is the percent yield?

Answer 1

Answer: The theoretical yield and percent yield of $[Co(NH_3)_4(H_2O)_2]Cl_2$ is 3.93 g and 30.53 % respectively

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

Given mass of $CoCl_2.6H_2O$ = 4.00 g

Molar mass of $CoCl_2.6H_2O$ = 238 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CoCl_2.6H_2O=\frac{4.00g}{238g/mol}=0.0168mol$

The chemical equation for the reaction of $CoCl_2.6H_2O$ to form  $[Co(NH_3)_4(H_2O)_2]Cl_2$ follows:

$CoCl_2.6H_2O+4NH_3\rightarrow [Co(NH_3)_4(H_2O)_2]Cl_2+4H_2O$

By Stoichiometry of the reaction:

1 mole of $CoCl_2.6H_2O$ produces 1 mole of $[Co(NH_3)_4(H_2O)_2]Cl_2$

So, 0.0168 moles of $CoCl_2.6H_2O$ will produce = $\frac{1}{1}\times 0.0168=0.0168mol$ of $[Co(NH_3)_4(H_2O)_2]Cl_2$

Now, calculating the mass of $[Co(NH_3)_4(H_2O)_2]Cl_2$ from equation 1, we get:

Molar mass of $[Co(NH_3)_4(H_2O)_2]Cl_2$ = 234 g/mol

Moles of $[Co(NH_3)_4(H_2O)_2]Cl_2$ = 0.0168 moles

Putting values in equation 1, we get:

$0.0168mol=\frac{\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2}{234g/mol}\\\\\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2=(0.0168mol\times 234g/mol)=3.93g$

To calculate the percentage yield of $[Co(NH_3)_4(H_2O)_2]Cl_2$, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $[Co(NH_3)_4(H_2O)_2]Cl_2$ = 1.20 g

Theoretical yield of $[Co(NH_3)_4(H_2O)_2]Cl_2$ = 3.93 g

Putting values in above equation, we get:

$\%\text{ yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=\frac{1.20g}{3.93g}\times 100\\\\\% \text{yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=30.53\%$

Hence, the theoretical yield and percent yield of $[Co(NH_3)_4(H_2O)_2]Cl_2$ is 3.93 g and 30.53 % respectively