Question

Question #a40d7

Answer 1

`64.4%`

Explanation:

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

`approx64.380%` rounded to three decimal places

Explanation:

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