\(a,n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)

PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

           0,075<-------------------------0,075

Cu không phản ứng với H₂SO₄ loãng

b, \(m_{Mg}=0,075.24=1,8\left(g\right)\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{1,8}{8}.100\%=22,5\%\\\%m_{Cu}=100\%-22,5\%=77,5\%\end{matrix}\right.\)

\(n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\ pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\) 
            0,075                                0,075 
         \(Cu+H_2SO_4-x->\) 
          \(m_{Mg}=0,075.24=1,6\left(g\right)\\ m_{Cu}=8-1,6=6,4\left(g\right)\) 
\(\%m_{Cu}=\dfrac{6,4}{8}.100\%=80\%\\ \%m_{Mg}=100-80\%=20\%\)

nH2= 0,35(mol)

a) PTHH: Mg +  2 HCl -> MgCl2 + H2

x_________2x_______x______x(mol)

PTHH: Fe + 2 HCl -> FeCl2 + H2

y________2y________y_____y(mol)

Ta có hpt: \(\left\{{}\begin{matrix}24x+56y=13,2\\x+y=0,35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,15\end{matrix}\right.\)

b) m=m(muối khan)= mMgCl2 + mFeCl2= 95.x+127y=95.0,2+127.0,15= 38,05(g)

a)

Gọi 

\(n_{Fe} = a(mol) ; n_{Mg} = b(mol)\\ \Rightarrow 56a + 24b = 13,2(1)\)

\(Mg + 2HCl \to MgCl_2 + H_2\\ Fe + 2HCl \to FeCl_2 + H_2\)

Theo PTHH : \(n_{H_2} = a + b = 0,35(mol)\)(2)

Từ (1)(2) suy ra a = 0,15 ;b = 0,2

Vậy : 

\(\%m_{Fe} = \dfrac{0,15.56}{13,2}.100\% = 63,64\%\\ \Rightarrow m_{Mg} = 100\% - 63,64\% = 36,36\%\)

b)

Ta có :\(n_{HCl} = 2n_{H_2} = 0,7(mol)\)

Bảo toàn khối lượng :

\(m_{muối} = m_{kim\ loại} + m_{HCl} - m_{H_2} = 13,2 + 0,7.36,5 - 0,35.2=38,05(gam)\)

Sửa đề: đktc → đkc

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

Ta có: 24n_Mg + 56n_Fe = 13,2 (1)

\(n_{H_2}=\dfrac{8,6765}{24,79}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=0,35\left(2\right)\)

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2.24}{13,2}.100\%\approx36,36\%\\\%m_{Fe}\approx63,64\%\end{matrix}\right.\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\)

⇒ m _{muối khan} = 0,2.95 + 0,15.127 = 38,05 (g)

Câu 1:

Gọi : nMg=a(mol); nMgO=b(mol) (a,b>0)

a) PTHH: Mg + 2 HCl -> MgCl2 + H2

a________2a_______a______a(mol)

MgO +2 HCl -> MgCl2 + H2O

b_____2b_______b___b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}24a+40b=8,8\\22,4a=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

=> mMg=0,2.24=4,8(g)

=>%mMg= (4,8/8,8).100=54,545%

=> %mMgO= 45,455%

b) m(muối)=mMg2+ + mCl- = 0,3. 24 + 0,6.35,5=28,5(g)

c) V=VddHCl=(2a+2b)/2=0,3(l)=300(ml)

Câu 2:

Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{HCl}=0,4\cdot2=0,8\left(mol\right)\end{matrix}\right.\)

PTHH: \(Ca+2HCl\rightarrow CaCl_2+H_2\uparrow\)

               0,2____0,4_____0,2____0,2   (mol)

           \(CaO+2HCl\rightarrow CaCl_2+H_2O\)

                0,2____0,4______0,2____0,2  (mol)

Ta có: \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,2\cdot40}{0,2\cdot40+0,2\cdot56}\cdot100\%\approx41,67\%\\\%m_{CaO}=58,33\%\\m_{CaCl_2}=\left(0,2+0,2\right)\cdot111=44,4\left(g\right)\end{matrix}\right.\)

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

b, Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{MgO}=y\left(mol\right)\end{matrix}\right.\)

⇒ 24x + 40y = 4,4 (1)

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

Theo PT: \(n_{H_2}=n_{Mg}=x\left(mol\right)\)

⇒ x = 0,1 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{4,4}.100\%\approx54,54\%\\\%m_{MgO}\approx45,46\%\end{matrix}\right.\)

c, Theo PT: \(\Sigma n_{HCl}=2n_{Mg}+2n_{MgO}=0,3\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,3}{2}=0,15\left(l\right)=150\left(ml\right)\)

Bạn tham khảo nhé!

\(a.Mg+2HCl->MgCl_2+H_2\\ MgO+2HCl->MgCl_2+H_2O\\ b.n_{H_2}=\dfrac{2,24}{22,4}=n_{Mg}=0,1mol\\ \%m_{Mg}=\dfrac{0,1.24}{6}=40\%;\%m_{MgO}=60\%\\ n_{MgO}=\dfrac{0,6.6}{40}=0,09\left(mol\right)\\ n_{MgCl_2}=0,1+0,09=0,19\left(mol\right)\\ n_{HCl}=0,19.2=0,38\left(mol\right)\\ V_{ddHCl}=\dfrac{0,38.36,5}{0,2.1,1}=63,0\left(mL\right)\\ C_{M\left(MgCl_2\right)}=\dfrac{0,19}{0,063}=3,0\left(M\right)\)