\(\begin{array}{l}
a)\\
nC{O_2} = \dfrac{{6,72 \times 40\% }}{{22,4}} = 0,12\,mol\\
nCO = 0,3 - 0,12 = 0,18mol\\
nC = nC{O_2} + nCO = 0,3\,mol\\
nO = 2nC{O_2} + nCO = 0,42\,mol\\
\% mC = \dfrac{{0,3 \times 12}}{{0,3 \times 12 + 0,42 \times 16}} \times 100\% = 34,88\% \\
\% mO = 100 - 34,88 = 65,12\% \\
b)\\
hh:MgO(a\,mol),CuO(b\,mol)\\
40a:80b = 1:2\\
= > a = b\\
= > 40a + 80a = 20 = > a = \frac{1}{6}\,mol\\
nMg = nCu = \dfrac{1}{6}\,mol\\
mO = \dfrac{1}{3}\,mol\\
\% mMg = 20\% \\
\% mCu = 53,33\% \\
\% mO = 26,67\%
\end{array}\)
