\(\begin{array}{l}
1)\\
MgC{O_3} + {H_2}S{O_4} \to MgS{O_4} + C{O_2} + {H_2}O\\
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
A:CuS{O_4},HCl,MgS{O_4}\\
B:BaS{O_4}\\
D:C{O_2}\\
2)\\
nC{O_2} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
 =  > nMgC{O_3} = nC{O_2} = 0,1\,mol\\
mMgC{O_3} = 0,1 \times 84 = 8,4g\\
nBaS{O_4} = \dfrac{{23,3}}{{233}} = 0,1\,mol\\
nBaC{l_2} = nBaS{O_4} = 0,1\,mol\\
mBaC{l_2} = 0,1 \times 208 = 20,8g\\
mCuO = 30 - 20,8 - 8,4 = 0,8g\\
3)\\
nCuO = \dfrac{{0,8}}{{80}} = 0,01\,mol\\
n{H_2}S{O_4} = (0,1 + 0,1 + 0,01) \times 116,3\%  = 0,24423\,mol\\
m{\rm{dd}}{H_2}S{O_4} = \dfrac{{0,24423 \times 98}}{{24,5\% }} = 97,692g\\
4)\\
m{\rm{dd}}spu = 30 + 97,692 - 0,1 \times 44 - 23,3 = 99,992g\\
C\% MgS{O_4} = \dfrac{{0,1 \times 120}}{{99,992}} \times 100\%  = 12\% \\
C\% HCl = \dfrac{{0,2 \times 36,5}}{{99,992}} \times 100\%  = 7,3\% \\
C\% CuS{O_4} = \dfrac{{0,01 \times 160}}{{99,992}} \times 100\%  = 1,6\% 
\end{array}\)