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

Step-by-step 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=(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 =
`(1)/(1)* 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* 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}}* 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=(1.20g)/(3.93g)* 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