Em tham khảo nha :
\(\begin{array}{l}
2)\\
a)\\
CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}O + C{O_2}\\
BaC{O_3} + 2HCl \to BaC{l_2} + {H_2}O + C{O_2}\\
{n_{HCl}} = 0,4 \times 1 = 0,4mol\\
hh:CaC{O_3}(a\,mol),BaC{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
2a + 2b = 0,4\\
111a + 208b = 31,9
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,1\\
{m_{CaC{O_3}}} = 0,1 \times 100 = 10g\\
{m_{BaC{O_3}}} = 0,1 \times 197 = 19,7g\\
m = 10 + 19,7 = 29,7g\\
b)\\
\% CaC{O_3} = \dfrac{{10}}{{29,7}} \times 100\% = 33,7\% \\
\% BaC{O_3} = 100 - 33,7 = 66,3\% \\
3)\\
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Fe = \dfrac{{5,6}}{{12}} \times 100\% = 46,7\% \\
\% Cu = 100 - 46,7 = 53,3\% \\
c)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,1mol\\
{m_{{H_2}S{O_4}}} = 0,1 \times 98 = 9,8g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{9,8}}{{200}} \times 100\% = 4,9\%
\end{array}\)
