Em tham khảo nha:
\(\begin{array}{l}
1)\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = \dfrac{5}{{100}} = 0,05\,mol\\
F{e_3}{O_4} + 4CO \xrightarrow{t^0} 3Fe + 4C{O_2}\\
CuO + CO \xrightarrow{t^0} Cu + C{O_2}\\
hh:F{e_3}{O_4}(a\,mol),CuO(b\,mol)\\
\left\{ \begin{array}{l}
4a + b = 0,05\\
56 \times 3a + 64b = 2,32
\end{array} \right.\\
\Rightarrow a = b = 0,01\,mol\\
{m_{F{e_3}{O_4}}} = 0,01 \times 232 = 2,32g\\
{m_{CuO}} = 0,01 \times 80 = 0,8g\\
2)\\
{m_O} = 15,2 - 13,6 = 1,6g\\
{n_{CO}} = {n_{C{O_2}}} = {n_O} = \frac{{1,6}}{{16}} = 0,1\,mol\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,1\,mol\\
{m_{CaC{O_3}}} = 0,1 \times 100 = 10g\\
{n_{CO}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
{n_{CO}} = 0,25 - 0,1 = 0,15\,mol\\
{M_{hh}} = \dfrac{{0,15 \times 28 + 0,1 \times 44}}{{0,25}} = 34,4\,g/mol\\
{d_{B/{H_2}}} = \frac{{34,4}}{2} = 17,2
\end{array}\)
