a. PTHH : Mg + 2HCl ➝ MgCl₂ + H₂ (1)

b. theo bài : nH₂ = 3,36 : 22,4 = 0,15 (mol)

theo (1) nMg = nH₂ = 0,15 (mol)

➞ mMg = 0,15 ✖ 24 = 3,6 (g)

➞ %mMg = (3,6 : 5)✖100 = 72%

➞ %mCu = 100% - 72% = 28%

c. theo (1) nHCl = 2nH₂ = 2✖0,15 = 0,3 (mol)

mHCl = 0,3✖36,5 = 10,95(g)

➜mddHCl = (10,95✖100):14,6 = 75(g)

d. dung dịch Y : MgCl₂

mdd_(spư)= 3,6+75-0,3 = 78,3(g)

theo (1) nMgCl₂ = nH₂ = 0,15(mol)

mMgCl₂ = 0,15✖95 = 14,25(g)

C%MgCl₂ = (14,25 : 78,3)✖100 = 18,199%

\(Zn+2HCl->ZnCl_2+H_2\\ Mg+2HCl->MgCl_2+H_2\\ n_{Zn}=a\\ n_{Mg}=b\\ 65a+24b=11,3g\\ n_{H_2}=a+b=\dfrac{6,72}{22,4}=0,3\\ a=0,1\\ m_{Zn}=65.0,1=6,5g\)

TN1: Gọi \(\left(n_{Mg};n_{Al};n_{Zn}\right)=\left(a;b;c\right)\)

=> 24a + 27b + 65c = 28,6 (1)

\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: 2Mg + O₂ --t^o--> 2MgO

             a--->0,5a

            4Al + 3O₂ --t^o--> 2Al₂O₃

             b-->0,75b

            2Zn + O₂ --t^o--> 2ZnO

             c--->0,5c

=> 0,5a + 0,75b + 0,5c = 0,5 (2)

TN2: Gọi \(\left(n_{Mg};n_{Al};n_{Zn}\right)=\left(ak;bk;ck\right)\)

=> ak + bk + ck = 0,8 (3)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

           ak----------------------->ak

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

           bk------------------------>1,5bk

           Zn + 2HCl --> ZnCl₂ + H₂

          ck---------------------->ck

=> \(ak+1,5bk+ck=\dfrac{22,4}{22,4}=1\) (4)

(1)(2)(3)(4) => \(\left\{{}\begin{matrix}a=0,2\\b=0,4\\c=0,2\\k=1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2.24}{28,6}.100\%=16,783\%\\\%m_{Al}=\dfrac{0,4.27}{28,6}.100\%=37,762\%\\\%m_{Zn}=\dfrac{0,2.65}{28,6}.100\%=45,455\end{matrix}\right.\)

\(n_{Cu} = a ; n_{Al} = b ; n_{Fe} = c(mol)\\ \Rightarrow 64a + 27b + 56c = 28,6(1)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = 1,5b + c = \dfrac{13,44}{22,4} = 0,6(2)\\ \text{Mặt khác} : n_{O_2} = \dfrac{8,96}{22,4} = 0,4(mol)\\ 2Cu + O_2 \xrightarrow{t^o} 2CuO\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 4Fe + 3O_2 \xrightarrow{t^o} 2Fe_2O_3\\ \)

Ta có :

\(\dfrac{n_X}{n_{O_2}}=\dfrac{a+b+c}{0,5a +0,75b + 0,75c} = \dfrac{0,6}{0,4}(3)\\ (1)(2)(3)\Rightarrow a = \dfrac{317}{1460} ; b = \dfrac{121}{365}; c = \dfrac{15}{146}\\ \%m_{Cu} = \dfrac{\dfrac{317}{1460}.64}{28,6}.100\% = 48,59\%\\ \%m_{Al} = \dfrac{\dfrac{121}{365}.27}{28,6}.100\% = 31,3\%\\ \%m_{Fe} = 100\% - 41,59\% - 31,3\% = 27,11\%\)

\(a,\) Đặt \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(g\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo đề ta có HPT: \(\left\{{}\begin{matrix}27x+24y=7,5\\1,5x+y=0,35\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,1\cdot27=2,7\left(g\right)\\m_{Mg}=7,5-2,7=4,8\left(g\right)\end{matrix}\right.\)

\(b,n_{HCl}=3n_{Al}+2n_{Mg}=0,3+0,4=0,7\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,7\cdot36,5=25,55\left(g\right)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{25,55\cdot100\%}{14,6\%}=175\left(g\right)\)

\(n_{HCl}=0,1mol\)

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

0,1      0,2         0,1          0,1

\(m_{FeCl_2}=0,1\cdot127=12,7g\)

\(V_{H_2}=0,1\cdot22,4=2,24l\)