TN1: Gọi (n_Cu, n_Al, n_Fe) = (a,b,c)

=> 64a + 27b + 56c = 14,3 (1)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             b----------------------->1,5b

            Fe + 2HCl --> FeCl₂ + H₂

              c----------------------->c

=> 1,5b + c = 0,3 (2)

TN2: Gọi (n_Cu, n_Al, n_Fe) = (ak,bk,ck)

=> ak + bk + ck = 0,6 (3)

\(n_{O_2}=\dfrac{44,8}{22,4}.20\%=0,4\left(mol\right)\)

PTHH: 2Cu + O₂ --t^o--> 2CuO

            ak--->0,5ak

            4Al + 3O₂ --t^o--> 2Al₂O₃

           bk--->0,75bk

            3Fe + 2O₂ --t^o--> Fe₃O₄

           ck-->\(\dfrac{2}{3}ck\)

=> 0,5ak + 0,75bk + \(\dfrac{2}{3}ck\) = 0,4 (4)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,05\left(mol\right)\\b=0,1\left(mol\right)\\c=0,15\left(mol\right)\\k=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,05.64}{14,3}.100\%=22,38\%\\\%m_{Al}=\dfrac{0,1.27}{14,3}.100\%=18,88\%\\\%m_{Fe}=\dfrac{0,15.56}{14,3}.100\%=58,74\%\end{matrix}\right.\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

            \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

            \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

a) Ta có: \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{12,7}{36,5}=\dfrac{127}{365}\left(mol\right)\\n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\end{matrix}\right.\)

Ta thấy: \(2n_{H_2}< n_{HCl}\) \(\Rightarrow\) Axit còn dư

b) Theo PTHH: \(n_{HCl\left(p/ứ\right)}=2n_{H_2}=0,3\left(mol\right)\) \(\Rightarrow m_{HCl}=0,3\cdot36,5=10,95\left(g\right)\)

Mặt khác: \(m_{H_2}=0,15\cdot2=0,3\left(g\right)\)

Bảo toàn khối lượng: \(m_{muối}=m_{KL}+m_{HCl\left(p/ứ\right)}-m_{H_2}=18,65\left(g\right)\)

Mg+H₂SO₄->MgSO₄+H₂

x-------------------------------x

2Al+3H₂SO₄->Al₂(SO₄)₃+3H₂

y-------------------------------------3\2y

Ta có :

\(\left\{{}\begin{matrix}24x+27y=11,7\\x+\dfrac{3}{2}y=0,6\end{matrix}\right.\)

=>x=0,15 mol , y=0,3 mol

=>m _MgSO4=0,15.120=18g

=>m _Al2(SO4)3=0,15.342=51,3g

Câu hỏi >?

a)

Mg + H₂SO₄ --> MgSO₄ + H₂

2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

Fe + H₂SO₄ --> FeSO₄ + H₂

b) 

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

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

            2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

            0,04-->0,06----------------->0,06

            Fe + H₂SO₄ --> FeSO₄ + H₂

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

=> a = n_H2SO4 = 0,1 + 0,06 + 0,15 = 0,31 (mol)

m = m_X - m_H2 = 0,1.24 + 0,04.27 + 0,15.56 - 2(0,1 + 0,06 + 0,15)

= 11,26 (g)

\(2Al+6HCl\to 2AlCl_3+3H_2\\ Fe+2HCl\to FeCl_2+H_2\\ n_{HCl}=0,2.4=0,8(mol)\\ \Rightarrow \begin{cases} 56.n_{Fe}+27.n_{Al}=22\\ 2.n_{Fe}+3.n_{Al}=0,8 \end{cases}\Rightarrow \begin{cases} n_{Fe}=0,39(mol)\\ n_{Al}=0,007(mol) \end{cases}\\ \Rightarrow \begin{cases} \%m_{Fe}=\dfrac{0,39.56}{22}.100\%=99,27\%\\ \%m_{Al}=100\%-99,27\%=0,73\% \end{cases}\)

Giả sử 21,4 gam X chứa \(\left\{{}\begin{matrix}Cu:a\left(mol\right)\\Fe:2a\left(mol\right)\\R:b\left(mol\right)\end{matrix}\right.\)

=> 176a + b.M_R = 21,4 (1)

\(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

            2a--------------------->2a

            2R + 2nHCl --> 2RCl_n + nH₂

              b---------------------->0,5bn

=> 2a + 0,5bn = 0,7 (2)

- 10,7g X chứa \(\left\{{}\begin{matrix}Cu:0,5a\left(mol\right)\\Fe:a\left(mol\right)\\R:0,5b\left(mol\right)\end{matrix}\right.\)

\(n_{Cl_2}=\dfrac{39,1-10,7}{71}=0,4\left(mol\right)\)

PTHH: Cu + Cl₂ --t^o--> CuCl₂

          0,5a->0,5a

            2Fe + 3Cl₂ --t^o--> 2FeCl₃

               a-->1,5a

           2R + nCl₂ --t^o--> 2RCl_n

          0,5b->0,25bn

=> 2a + 0,25bn = 0,4 (3)

(2)(3) => a = 0,05 (mol); bn = 1,2

=> \(\left\{{}\begin{matrix}n_{Cu}=0,05\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,05.64}{21,4}.100\%=14,95\%\\\%m_{Fe}=\dfrac{0,1.56}{21,4}.100\%=26,17\%\end{matrix}\right.\)

a)

$4Na + O_2 \xrightarrow{t^o} 2Na_2O$

$2Mg + O_2  \xrightarrow{t^o} 2MgO$
$4Al + 3O_2  \xrightarrow{t^o} 2Al_2O_3$

$2Na + 2HCl \to 2NaCl + H_2$
$Mg + 2HCl \to MgCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$

b)

Bảo toàn khối lượng : $m_{O_2} = 4,08 - 2,48 = 1,6(gam)$
$n_{O_2} = \dfrac{1,6}{32} = 0,05(mol)$

Đốt 2,48 gam X cần 0,05 mol $O_2$
Suy ra, đốt 4,96 gam X cần 0,1 mol $O_2$

Mà : \(\dfrac{1}{4}n_{Na}+\dfrac{1}{2}n_{Mg}+\dfrac{3}{4}n_{Al}=n_{O_2}=0,1\)

Theo PTHH : 

\(n_{H_2}=\dfrac{1}{2}n_{Na}+n_{Mg}+\dfrac{3}{2}n_{Al}=2\left(\dfrac{1}{4}n_{Na}+\dfrac{1}{2}n_{Mg}+\dfrac{3}{4}n_{Al}\right)=2.0,1=0,2\)$V = 0,2.22,4 = 4,48(lít)$
$n_{HCl} = 2n_{H_2} = 0,4(mol)$
Bảo toàn khối lượng : $m = 4,96 + 0,4.36,5 - 0,2.2 = 19,16(gam)$

nHCl = 1 (mol)

Mg + 2HCl ---> MgCl2 + H2

Fe + 2HCl ---> FeCl2 + H2

2Al + 6HCl ---> 2AlCl3 + 3H2

Theo PTHH, ta thấy: \(n_{H_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.1=0,5\left(mol\right)\)

\(\Rightarrow m_{H_2}=0,5.2=1\left(g\right)\)

Bảo toàn khối lượng => m = 13,4 + 36,5 - 1 = 48,9 (g)

nHCl = 1 (mol)

Mg + 2HCl ---> MgCl2 + H2

Fe + 2HCl ---> FeCl2 + H2

2Al + 6HCl ---> 2AlCl3 + 3H2

Theo PTHH, ta thấy: nH2=12.nHCl=12.1=0,5(mol)nH2=12.nHCl=12.1=0,5(mol)

⇒mH2=0,5.2=1(g)⇒mH2=0,5.2=1(g)

Bảo toàn khối lượng => m = 13,4 + 36,5 - 1 = 48,9 (g)