Question a40d7

Feb 24, 2016

64.4%

Explanation:

The chemical equation is:

$C u C {O}_{3} \to C u O + C {O}_{2}$

You see that for each CuCO_3 you form one CuO.

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

$M \left(C u O\right) = 63.55 + 16.000 = 79.55 g / m o l$

$M \left(C u C {O}_{3}\right) = 63.55 + 12.011 + 3 \times 16.000 = 123.56 g / m o l$

The theorical percentage of CuO formed is:

100xx79.55/123.56=64.4%

Feb 24, 2016

approx64.380% rounded to three decimal places

Explanation:

The chemical equation of the given reaction is

${\text{CuCO"_3 ->"CuO"+"CO}}_{2}$

We see that for each molecule of ${\text{CuCO}}_{3}$ one molecule of $\text{CuO}$ is formed.

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

$M \left(\text{CuO}\right) = 63.546 + 15.9994 = 79.5454$

$M \left({\text{CuCO}}_{3}\right) = 63.546 + 12.011 + 3 \times 15.9994 = 123.5552$

Theoretical percentage of $\text{CuO}$ formed is

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