\(Fe\left(OH\right)_2+2HCl\rightarrow FeCl_2+2H_2O\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{HCl}=\dfrac{25,55}{35,6}=0,7\left(mol\right)\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,7}{3}\), ta được HCl dư.

Theo PT: \(n_{HCl\left(pư\right)}=3n_{Al}=0,6\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,7-0,6=0,1\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

\(Fe\left(OH\right)_2+H_2SO_4\rightarrow FeSO_4+2H_2O\)

\(n_{HCl}=\dfrac{100.16,79\%}{36,5}=0,46\left(mol\right)\)

Gọi: CTHH oxit cần tìm là X₂O

\(\Rightarrow n_{X_2O}=\dfrac{29,14}{2M_X+16}\left(mol\right)\)

BTNT Cl, có: \(n_{XCl}=n_{HCl}=0,46\left(mol\right)\)

BTNT X, có: \(2n_{X_2O}=n_{XCl}+n_{XOH}\)

\(\Rightarrow n_{XOH}=\dfrac{2.29,14}{2M_X+16}-0,46\left(mol\right)\)

Mà: m_XCl + m_XOH = 46,11

\(\Rightarrow0,46.\left(M_X+35,5\right)+\left(\dfrac{2.29,14}{2M_X+16}-0,46\right).\left(M_X+17\right)=46,11\)

\(\Rightarrow M_X=23\left(g/mol\right)\)

→ X là Na

Vậy: CTHH cần tìm là Na₂O

cho mình hỏi là vì sao ở nxoh  phải trừ 0,46 vậy

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)

PT: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

Theo PT: \(n_{Na_2SO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)

\(n_{NaOH}=2n_{H_2SO_4}=0,4\left(mol\right)\)

\(\Rightarrow V_{ddNaOH}=\dfrac{0,4}{1}=0,4\left(l\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,2}{0,2+0,4}=\dfrac{1}{3}\left(M\right)\)

\(n_{K_2O}=\dfrac{23,5}{94}=0,25\left(mol\right)\)

PT: \(K_2O+H_2O\rightarrow2KOH\)

Theo PT: \(n_{KOH}=2n_{K_2O}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Gọi: CTHH của oxit là Fe_xO_y

Ta có: m_O = 7,2 - m_Fe = 1,6 (g) \(\Rightarrow n_O=\dfrac{1,6}{16}=0,1\left(mol\right)\)

⇒ x:y = 0,1:0,1 = 1:1

→ CTHH cần tìm là FeO.

\(n_{NaOH}=0,025.8=0,2\left(mol\right)\)

\(n_{HCl}=0,05.6=0,3\left(mol\right)\)

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\), ta được HCl dư.

→ Dung dịch sau pư làm quỳ hóa đỏ.

\(a,n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{8}{80}=0,1\left(mol\right)\\ b,n_{Fe_2\left(SO_4\right)_3}=\dfrac{m_{Fe_2\left(SO_4\right)_3}}{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{20}{400}=0,05\left(mol\right)\)

\(a.n_{CuO}=\dfrac{8}{80}=0,1mol\\ b.n_{Fe_2\left(SO_4\right)_3}=\dfrac{20}{400}=0,05mol\)

\(CaCO_3\xrightarrow[t^0]{}CaO+CO_2\\ \Rightarrow\dfrac{1}{100}=\dfrac{m_{CaO\left(LT\right)}}{56}\\ \Rightarrow m_{CaO\left(LT\right)}=0,56\left(tấn\right)\\ m_{CaO\left(TT\right)}=0,56.90\%=0,504\left(tấn\right)\)