Bài 22. Luyện tập chương II: Kim loại

\(a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ b,n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{Al}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ c,n_{H_2SO_4}=0,3(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,3.98}{9,8\%}=300(g)\\ d,n_{Al_2(SO_4)_3}=0,1(mol)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{5,4+300-0,3.2}.100\%=11,22\%\)

mO2=10,2−5,4=4,8g

=>nO2=0,15mol

=>VO2=0,15.22,4=3,36 l()

4M+3O2−>2M2O3

0,2<−0,15                (mol)

MM=5,4x0,2=27 g/mol

Vậy M là nhôm (Al)

\(4M+3O_2-t^0>2M_2O_3\)

5,4-------------------10,2

4M--------------------2(2M+16.3)

=> 5,4.2(2M+16.3)=4M.10,2

Giải PT trên => M=27 ( Al )

ADĐLBTKL ta có 
\(m_{O_2}=m_{M_2O_3}-m_M=10,2-5,4=4,8\left(g\right)\\ n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\\ 4M+3O_2\underrightarrow{t^o}2M_2O_3\) 
0,2       0,15 
\(M_M=\dfrac{5,4}{0,2}=27\left(\dfrac{g}{mol}\right)\) 
Mà sp là M₂O₃ => M hóa trị III 
=> M là Al (nhôm) 

\(A.2A+3Cl_2\rightarrow2ACl_3\\ n_A:2=n_{ACl_3}:2\\ \Leftrightarrow\dfrac{10,8}{A}:2=\dfrac{53,4}{A+106,5}:2\\ \Leftrightarrow A=27,Al\\ n_{Cl_2}=1,5n_{Al}=1,5\cdot\dfrac{10,8}{27}=0,6mol\\ V_{Cl_2}=0,6.22,4=13,44l\)

a) Gọi \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

             a--->a---------->a-------->a

            Fe + H₂SO₄ --> FeSO₄ + H₂

             b--->b----------->b------>b

=> \(m_{H_2SO_4}=98a+98b\left(g\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{\left(98a+98b\right).100}{19,6}=500a+500b\left(g\right)\)

m_{dd sau pư} = 24a + 56b + 500a + 500b - 2a - 2b = 522a + 554b (g)

Có: \(C\%_{FeSO_4}=\dfrac{152b}{522a+554b}.100\%=7,17\%\)

=> a = 3b

\(C\%_{MgSO_4}=\dfrac{120a}{522a+554b}.100\%=16,98\%\)

b)

Có: \(\left\{{}\begin{matrix}a=3b\\24a+56b=1,92\end{matrix}\right.\)

=> a = 0,045; b = 0,015

\(n_{CuSO_4}=0,1.1=0,1\left(mol\right)\)

PTHH: Mg + CuSO₄ --> MgSO₄ + Cu

         0,045->0,045----->0,045

            Fe + CuSO₄ --> FeSO₄ + Cu

       0,015-->0,015----->0,015

=> \(\left\{{}\begin{matrix}n_{CuSO_4\left(dư\right)}=0,04\left(mol\right)\\n_{MgSO_4}=0,045\left(mol\right)\\n_{FeSO_4}=0,015\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C_{M\left(CuSO_4\left(dư\right)\right)}=\dfrac{0,04}{0,1}=0,4M\\C_{M\left(MgSO_4\right)}=\dfrac{0,045}{0,1}=0,45M\\C_{M\left(FeSO_4\right)}=\dfrac{0,015}{0,1}=0,15M\end{matrix}\right.\)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ Ba\left(OH\right)_2+2Zn+6H_2O\rightarrow Ba\left(Zn\left(OH\right)_4\right)_2+3H_2\)
\(\Rightarrow\)Dung dịch Y gồm \(Ba\left(OH\right)_2\left(dư\right)\) và \(Ba\left(Zn\left(OH\right)_4\right)_2\), khí Z là \(H_2\), chất rắn A gồm \(MgO,Fe_2O_3,Cu\)
\(H_2+MgO\rightarrow Mg+H_2O\\ Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
\(\Rightarrow\)B gồm \(Mg,Fe,Cu\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\)
\(\Rightarrow\)T gồm \(MgCl_2,FeCl_2,HCl\left(dư\right)\), C là Cu
\(FeCl_2+2HCl+4AgNO_3\rightarrow Fe\left(NO_3\right)_3+4AgCl+NO_2+H_2O\\ MgCl_2+2AgNO_3\rightarrow Mg\left(NO_3\right)_2+2AgCl\)
\(AgNO_3+HCl\rightarrow AgCl+HNO_3\)
\(\Rightarrow\)Khí hóa nâu trong không khí là NO₂, E gồm AgCl
_{P/s: mình nghĩ chất rắn E chỉ có AgCl chứ còn chất nào kết tủa nữa đâu?:))}

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

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

c, Ta có: 65n_Zn + 24n_Mg = 10,1 (1)

Theo PT: \(n_{H_2}=n_{Zn}+n_{Mg}=0,25\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,1\left(mol\right)\\n_{Mg}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,1.65=6,5\left(g\right)\\m_{Mg}=0,15.24=3,6\left(g\right)\end{matrix}\right.\)

\(A.Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ B.n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Zn}=a,n_{Mg}=b\\ \Rightarrow\left\{{}\begin{matrix}65a+24b=10,1\\a+b=0,25\end{matrix}\right.\\ \Rightarrow a=0,1;b=0,15\\ n_{HCl}=0,1.2+0,15.2=0,5mol\\ C_{M_{HCl}}=\dfrac{0,5}{0,1}=5M\\ C.m_{Zn}=0,1.65=6,5g\\ m_{Mg}=10,1-6,5=3,6g\)

\(a,PTHH:X+O_2\xrightarrow{t^o}XO_2\\ b,n_{X}=n_{XO_2}\\ \Rightarrow \dfrac{2}{M_X}=\dfrac{2,54}{M_X+32}\\ \Rightarrow M_X\approx 118,5(g/mol)\)

Vậy X là thiếc (Sn)