a,\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:    0,05     0,1        0,05     0,05

PTHH: MgO + 2HCl → MgCl₂ + H₂O

Mol:       0,1      0,2          0,1

⇒ m_Mg = 0,05.24 = 1,2 (g)

    m_MgO = 5,2 - 1,2 = 4 (g)

b,\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

⇒ n_{HCl đã dùng} = 0,1+0,2 = 0,3 (mol)

\(C_{M_{ddHCl}}=\dfrac{0,3}{0,5}=0,6\left(l\right)\)

c,\(C_{M_{MgCl_2}}=\dfrac{0,05+0,1}{0,6}=0,25M\) 

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

           0,2<--0,4<------0,2<-----0,2

=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,2.65}{21,1}.100\%=61,61\%\\\%m_{ZnO}=100\%-61,61\%=38,39\%\end{matrix}\right.\)

\(n_{ZnO}=\dfrac{21,1-0,2.65}{81}=0,1\left(mol\right)\)

PTHH: ZnO + 2HCl ---> ZnCl₂ + H₂O

            0,1---->0,2------>0,1

=> \(C\%_{HCl}=\dfrac{\left(0,2+0,4\right).36,5}{200}.100\%=10,95\%\)

\(m_{mu\text{ố}i}=m_{ZnCl_2}=\left(0,1+0,2\right).136=40,8\left(g\right)\)

\(n_O=\dfrac{21,2-14}{16}=0,45\left(mol\right)\)

=> n_H2O = 0,45 (mol)

Giả sử có V (lít) dd Z

=> \(\left\{{}\begin{matrix}n_{HCl}=0,1V\left(mol\right)\\n_{H_2SO_4}=0,1V\left(mol\right)\end{matrix}\right.\)

Bảo toàn H: 0,1V + 2.0,1V = 0,45.2

=> V = 3 (lít)

TN1: Gọi (n_Al; n_Zn; n_Fe) = (a; b; c)

=>27a + 65b + 56c = 20,4 (1)

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

              a----------------------->1,5a

             Zn + 2HCl --> ZnCl₂ + H₂

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

              Fe + 2HCl --> FeCl₂ + H₂

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

=> \(1,5a+b+c=0,45\) (2)

TN2: Gọi (n_Al; n_Zn; n_Fe) = (ak; bk; ck)

=> ak + bk + ck = 0,2 (3)

\(n_{Cl_2}=\dfrac{6,16}{22,4}=0,275\left(mol\right)\)

PTHH: 2Al + 3Cl₂ --t^o--> 2AlCl₃

            ak-->1,5ak

             Zn + Cl₂ -t^o-> ZnCl₂

            bk--->bk

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

              ck--->1,5ck

=> 1,5ak + bk + 1,5ck = 0,275 (4)

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

=> \(\%Fe=\dfrac{0,2.56}{20,4}.100\%=54,9\%\)

a) Sửa đề: dd H₂SO₄ 9,8%

Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2}=0,35\cdot2=0,7\left(g\right)\)

Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{H_2}=0,35\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,35\cdot98}{9,8\%}=350\left(g\right)\)

\(\Rightarrow m_{dd}=m_{KL}+m_{H_2SO_4}-m_{H_2}=361,6\left(g\right)\)

b) Tương tự câu a

câu 1.  giải hệ phương trình: \(\begin{cases}27x+65y=6,05\\133,5x+136y=13,15\end{cases}\)=>\(\begin{cases}x=\frac{3}{470}\\y=\frac{17}{188}\end{cases}\)=> nHCL= 3.(3/470)+2.(17/188)=0,2(MOL)

==> mct= 7,3 (g)

ta có C%=(mct/mdd).100%===> mdd=73(g)

câu 2.

từ ct trên===> mct của AgN03=17(g)===> nAgN03=0,1(mo)====> C_M HCl= n/V=2/3(M)

câu 3.

ta có Al+ 3HCl===> AlCl₃+ 3/2H₂

        _{0.2(mol)======>               3/2. 0,2=0.3( mol)}

_=====> v H2= 0,3. 22,4=6,72(l)

câu 4.  CaO+ 2H20=====> Ca(OH)2+ H2

2NaCl======>2 Na+ Cl2( điện phân nóng chảy)

CaOH)2+ Cl2=====> CaOCl2+ H20

                                    ( clorua vôi)

Zn+2HCl→ZnCl2+H2

 2Al+6HCl→2AlCl3+3H2

Gọi số mol HCllà x

n H2=\(\dfrac{1}{2}\)n HCl=0,5x

Bảo toàn khối lượng: mkl+mHCl=mmuối+mH2

6,05+36,5x=13,15+0,5x.2

→x=0,2 mol

mHCl=0,2.36,5=7,3 gam

=>m =mddHCl=\(\dfrac{7,3}{10\%}\)=73gam

Gọi : \(\left\{{}\begin{matrix}n_{Al_2O_3}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\)⇒ 102a + 65b = 2,505(1)

\(Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O\\ Zn + 2HCl \to ZnCl_2 + H_2\)

Muối gồm : \(\left\{{}\begin{matrix}AlCl_3:2a\left(mol\right)\\ZnCl_2:b\left(mol\right)\end{matrix}\right.\)⇒ 133,5.2a + 136b = 6,045(2)

Từ (1)(2) suy ra : a = 0,015 ; b = 0,015

Vậy :

\(\%m_{Al_2O_3} = \dfrac{0,015.102}{2,505}.100\% = 61,08\%\\ \%m_{Zn} = 100\% - 61,08\% = 38,92\%\)

Theo PTHH : \(n_{HCl} = 6a + 2b = 0,12(mol)\\ \Rightarrow C\%_{HCl} = \dfrac{0,12.36,5}{200}.100\% = 2,19\%\)