a,
$nH_{2} =\frac{6,72}{22,4}=0,3 \\ Al^{0} \to Al^{3+}+3e \\ H_{2}^{+1} +2e\to H_{2}^{0} \\ ⇒\text{BTe ta có: } 3nAl=2nH_{2}⇒nAl=0,2 \\nSO_{2}=\frac{13,44}{22,4}=0,6 \\ Al^{0} \to Al^{3+}+3e\\ Cu^{0} \to Cu^{2+}+2e \\ S^{+6} +2e \to S^{+4} \\ ⇒\text{BTe ta có: } 2nCu+3nAl=2nSO_{2} ⇒nCu=0,3 \\ \%mCu=\frac{0,3.64}{0,3.64+0,2.27}.100=78,05\% \\ \%mAl=100-78,05=21,95\%$
b,
$2Al+6H_{2}SO_{4} \to Al_{2}(SO_{4})_{3}+3SO_{2}+6H_{2}O \\ Cu+2H_{2}SO_{4} \to CuSO_{4}+SO_{2}+2H_{2}O \\ nH_{2}SO_{4}\text{phản ứng}=2nSO_{2}=0,6.2=1,2 \\ nH_{2}SO_{4}\text{ban đầu}=1,2.120\%=1,44 \\nH_{2}SO_{4} dư=1,44-1,2=0,24 \\ \text{Ta có :} nCu=nCuSO_{4} ; nAl=2nAl_{2}(SO_{4})_{3} ⇒nAl_{2}(SO_{4})_{3}=0,1 \\ mdd_{sau}=200+0,3.64+0,2.27-0,6.64=186,2g \\ C\%H_{2}SO_{4}dư=\frac{0,24.98}{186,2}.100=12,63\% \\ C\%CuSO_{4}=\frac{0,3.160}{186,2}.100=25,78\% \\ C\%Al_{2}(SO_{4})_{3}=\frac{0,1.342}{186,2}.100=18,37\%$
c,
$Ba(OH)_{2} +H_{2}SO_{4} \to BaSO_{4}+H_{2}O \\ 3Ba(OH)_{2} +Al_{2}(SO_{4})_{3} \to 2Al(OH)_{3}+3BaSO_{4} \\ 2Al(OH)_{3}+Ba(OH)_{2} \to 4H_{2}O+Ba(AlO_{2})_{2} \\ CuSO_{4}+Ba(OH)_{2} \to BaSO_{4}+Cu(OH)_{2} \\ Cu(OH)_{2} \to CuO+H_{2}O \\ nBaSO_{4}=nH_{2}SO_{4}dư+3nAl_{2}(SO_{4})_{3}+nCuSO_{4} =0,24+0,1.3+0,3=0,84 \\ nCuO=nCu(OH)_{2}=nCuSO_{4}=0,3 \\ m_{\text{chất rắn}}=mBaSO_{4}+mCuO=0,84.233+0,3.80=219,72g$