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Answer 1 · 2016-02-24T12:32:13

#64.4%#

The chemical equation is:

#CuCO_3 ->CuO+CO_2#

You see that for each CuCO_3 you form one CuO.

So, you have only to compare the relative molecular masses of both compounds :

#M(CuO)=63.55+16.000=79.55 g//mol#

#M(CuCO_3)=63.55+12.011+3xx16.000=123.56 g//mol#

The theorical percentage of CuO formed is:

#100xx79.55/123.56=64.4%#

Answer 2 · 2016-02-24T12:54:33

#approx64.380%# rounded to three decimal places

The chemical equation of the given reaction is

#"CuCO"_3 ->"CuO"+"CO"_2#

We see that for each molecule of #"CuCO"_3# one molecule of #"CuO"# is formed.

Therefore we need to compare the relative molecular masses of these two compounds.

#M("CuO")=63.546+15.9994=79.5454 #

#M("CuCO"_3)=63.546+12.011+3xx15.9994=123.5552#

Theoretical percentage of #"CuO"# formed is

#79.5454/123.5552xx100approx64.380%# rounded to three decimal places