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?

The theoretical yield and percent yield of is 3.93 g and 30.53 % respectively

Explanation:

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

    .....(1)

Given mass of = 4.00 g

Molar mass of = 238 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of to form   follows:

By Stoichiometry of the reaction:

1 mole of produces 1 mole of

So, 0.0168 moles of will produce = of

Now, calculating the mass of from equation 1, we get:

Molar mass of = 234 g/mol

Moles of = 0.0168 moles

Putting values in equation 1, we get:

To calculate the percentage yield of , we use the equation:

Experimental yield of = 1.20 g

Theoretical yield of = 3.93 g

Putting values in above equation, we get:

Hence, the theoretical yield and percent yield of is 3.93 g and 30.53 % respectively

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