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Question- I was doing an experiment and I started off with 12 grams of (CoCl2-6H20) and I produced 6.0282 grams of [Co(NH3)6]Cl3. So for my present yield I did

12.0 G Cocl2/ 129.839 mol Cocl2=0.09242 mol C02+ used

6.0282 g [Co(NH3)6]Cl3/ 215.5856 g/ mol [Co(NH3)6]Cl3= 0.02796 Co3+ recovered

% yield = 0.02796/ 0.09242 *100= 30.3 % is this correct?

As well for converting the CoCl2 to mol must I utilize the molar weight of CoCl2-6H2O or just the molar weight of CoCl2?

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Category:- Chemistry
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