\(\text{Chọn C}\\ H_2SO_4đ \text{ Ko tác dụng}: Au, Pt\)

này dùng bt nha 

Chưa xét Al(OH)3

\(n_{Ba(OH)_2}=0,25(mol)\\ \to n_{Ba^{2+}}=0,25(mol); n_{OH^{-}}=0,25.2=0,5(mol)\\ n_{Al_2(SO_4)_4}=\frac{25,65}{342}=0,075(mol)\\ \to n_{Al^{3+}}=0,075.2=0,15(mol); n_{SO_4^{2-}}=0,075.3=0,225(mol)\\ Ba^{2+} +SO_4^{2-} \to BaSO_4\\ 0,25>0,225\\ \Leftrightarrow Ba^{2+} >SO_4^{2-}\\ \to n_{BaSO_4}=0,225(mol)\\ m_{BaSO_4}=0,225.233=52,425(mol)\\ Al^{3+} + 3OH^{-} \to Al(OH)_3\\ 0,15<0,5\\ \Leftrightarrow Al^{3+} < OH^{-} n_{Al(OH)_3}=0,15(mol) \\ Al(OH)_3+ OH^{-} \to AlO_2^{-} +2H_2O\\ n_{OH^{-}du}=0,5-0,15.3=0,05(mol)\\ \to n_{Al(OH_3}=0,15-0,05=0,1(mol)\\ m_{Al(OH)_3}=0,1.78=7,8(g)\\ \to m_{\downarrow}=52,425+7,8=60,225(mol)\\ \to D \)

\(Mg+2H_2SO_4\buildrel{{t^o}}\over\longrightarrow MgSO_4+SO_2+2H_2O\\ Ca(OH)_2+SO_2 \to CaSO_3+H_2O n_{Mg}=\frac{12}{24}=0,5(mol)\\ n_{SO_2}=n_{Mg}=0,5(mol)\\ n_{CaSO_3}=n_{SO_2}=0,5(mol)\\ m_{CaSO_3}=0,5.120=60(g)\\ \to D \)

\(2FeSO_4+2H_2SO_4 \buildrel{{t^o}}\over\longrightarrow Fe_2(SO_4)_3+SO_2+2H_2O\\ n_{FeSO_4}=\frac{15,2}{152}=0,1(mol)\\ n_{H_2SO_4}=n_{FeSO_4}=0,1(mol)\\ \to A \)

\(\text{Chọn A}\\ BaCl_2+H_2SO_4 \to BaSO_4+2HCl\)

\(\text{Chọn C}\\ 2Ag+2H_2SO_4 \buildrel{{t^o}}\over\longrightarrow Ag_2SO_4+SO_2+2H_2O\)

\(\text{Chọn D}\\\ Cu+2H_2SO_4\buildrel{{t^o}}\over\longrightarrow CuSO_4+SO_2+2H_2O\)

\(\text{Chọn D vì tạo sủi bọt khí H2}\\ Fe+H_2SO_4 \to FeSO_4+H_2\)