a. \(PTHH:3H_2+Fe_2O_3\rightarrow2Fe+3H_2O\)

b. \(n_{Fe_2O_3}=\dfrac{m_{Fe_2O_3}}{M_{Fe_2O_3}}=\dfrac{40}{160}=0,25\left(mol\right)\)

- Mol theo PTHH : \(3:1:2:3\)

- Mol theo phản ứng : \(0,75\leftarrow0,25\rightarrow0,5\rightarrow0,75\)

\(\Rightarrow n_{Fe}=0,5\left(mol\right)\)

\(\Rightarrow m_{Fe}=n_{Fe}.M_{Fe}=0,5.56=28\left(g\right)\)

c. Ta có : \(n_{Fe_2O_3}=0,25\left(mol\right);n_{H_2}=0,3\left(mol\right)\)

Do \(0,25< 0,3\) ⇒ H₂ dư.

a

\(2Fe+6H_2SO_{4.đặc}\underrightarrow{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

b

\(2Na+2HCl\rightarrow2NaCl+H_2\)

c

\(8Al+30HNO_3\rightarrow8Al\left(NO_3\right)_3+3NH_4NO_3+9H_2O\)

d

\(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)

\(Ba\left(OH\right)_2+\left(NH_4\right)_2SO_4\rightarrow BaSO_4+2NH_3+2H_2O\)

e

\(2AlCl_3+3Na_2CO_3+H_2O\rightarrow2Al\left(OH\right)_3+6NaCl+3CO_2\)

f

\(HCl+NaAlO_2+H_2O\rightarrow Al\left(OH\right)_3+NaCl\)

a: \(2Fe+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+3H_2O\)(H2SO4 đặc nóng)

b: \(2Na+2HCl\rightarrow2NaCl+H_2\uparrow\)

c: \(8Al+30HNO_3\rightarrow8Al\left(NO_3\right)_3+3NH_4NO_3+9H_2O\)

d; \(Ba+\left(NH_4\right)_2SO_4\rightarrow BaSO_4\downarrow+2NH_3+H_2O\)

e: \(2AlCl_3+3Na_2CO_3+3H_2O\rightarrow6NaCl+2Al\left(OH\right)_3\downarrow+3CO_2\uparrow\)

f: \(HCl+NaAlO_2+H_2O\rightarrow Al\left(OH\right)_3+NaCl\)

a.b.\(n_{Fe_2O_3}=\dfrac{m_{Fe_2O_3}}{M_{Fe_2O_3}}=\dfrac{16}{160}=0,1mol\)

\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{13,44}{22,4}=0,6mol\)

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

0,1    <     0,6                                     ( mol )

0,1            0,3               0,2                        ( mol )

\(m_{Fe}=n_{Fe}.M_{Fe}=0,2.56=11,2g\)

c.\(n_{H_2}=0,6-0,3=0,3mol\)

\(CuO+H_2\rightarrow Cu+H_2O\)

 0,3      0,3                         ( mol )

\(m_{CuO}=n_{CuO}.M_{CuO}=0,3.80=24g\)

a) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Fe + H₂SO_4(l) ---> FeSO₄ + H₂

            0,1<-------------------------------0,1

\(m_{Fe}=0,1.56=5,6\left(g\right)\\ n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH:

2Fe + 6H₂SO_4(đ,n) ---> Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

0,1-------------------------------------------->0,15

2Ag + 2H₂SO_4(đ,n) ---> Ag₂SO₄ + SO₂ + 2H₂O

0,2<----------------------------------------0,1

=> m_Ag = 0,2.108 = 21,6 (g)

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{5,6+21,6}.100\%=20,44\%\\\%m_{Ag}=100\%-20,44\%=79,56\%\end{matrix}\right.\)

b) n_NaOH = 0,25.1,5 = 0,375 (mol)

\(T=\dfrac{0,375}{0,25}=1,5\) => Tạo cả 2 muối

PTHH:

2NaOH + SO₂ ---> Na₂SO₃ + H₂O

0,375--->0,1875--->0,1875

Na₂SO₃ + SO₂ + H₂O ---> 2NaHSO₃

0,0625<---0,0625----------->0,125

=> \(\left\{{}\begin{matrix}C_{M\left(Na_2SO_3\right)}=\dfrac{0,1875-0,0625}{0,25}=0,48M\\C_{M\left(NaHSO_3\right)}=\dfrac{0,125}{0,25}=0,5M\end{matrix}\right.\)

Gọi \(\left\{{}\begin{matrix}n_{RO}=2x\\n_{Fe_2O_3}=x\end{matrix}\right.\)

Theo đề:

\(R.2x+56.2x=31,56\)

\(\Rightarrow2x.R=31,56-112x\)

Mặt khác: \(2x.\left(R+16\right)+160x=36,36\)

\(\Leftrightarrow2xR+32x+160x=36,6\Leftrightarrow31,56-112x+32x+160x=36,36\)

=>  x = 0,06

\(\Rightarrow R=\dfrac{31,56-112.0,06}{2.0,06}=207\)

Vậy kim loại R là Pb (chì).