PTHH :

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

0,1      0,2           0,1         0,1 

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

0,09      0,18          0,09      0,09 

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

\(a,\%m_{Mg}=\dfrac{0,1.24}{6}=40\%\)

\(\%m_{MgO}=100\%-40\%=60\%\)

\(n_{MgO}=\dfrac{6-2,4}{40}=0,09\left(mol\right)\)

\(b,m_{HCl}=\left(0,2+0,18\right).36,5=13,87\left(g\right)\)

\(m_{ddHCl}=\dfrac{13,87.100}{20}=69,35\left(g\right)\)

\(V_{ddHCl}=\dfrac{m}{D}=\dfrac{69,35}{1,1}\approx63\left(ml\right)\) ( cái này mình nghĩ đề phải là D bạn nhé tại vì khối lượng riêng của HCl là 1,18g/ml ) 

\(c,m_{MgCl_2}=\left(0,09+0,1\right).95=18,05\left(g\right)\)

\(m_{ddMgCl_2}=6+69,35-0,2=75,15\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{18,05}{75,15}.100\%\approx24,02\%\)

Cứu..........😭😭😭😭😭😭😭😭

\(a.n_{Mg}=n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ \%m_{Mg}=\dfrac{0,1.24}{6}=40\%\\ \%m_{MgO}=60\%\\ b.n_{MgO}=\dfrac{3,6}{40}=0,09mol\\ V_{ddHCl}=\dfrac{36,5\cdot2\cdot0,19}{0,2\cdot1,1}=63,05mL\\ c.m_{ddsau}=6+63,05\cdot1,1=75,35g\\ C\%_{MgCl_2}=\dfrac{95\cdot0,19}{75,35}=23,95\%\)

nH2=\(\dfrac{5,6}{22,4}\)=0,25(mol)=nFeCl2nH2=5,622,4=0,25(mol)=nFeCl2

⇒mFeCl2=0,25⋅127=31,75(g)⇒mFeCl2=0,25⋅127=31,75(g)

c) Theo PTHH: nH2=nFe=0,25molnH2=nFe=0,25mol ⇒mFe=0,25⋅56=14(g)

⇒mFe=0,25⋅56=14(g)

⇒%mF

a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

b) Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)=n_{FeCl_2}\) 

\(\Rightarrow m_{FeCl_2}=0,25\cdot127=31,75\left(g\right)\)

c) Theo PTHH: \(n_{H_2}=n_{Fe}=0,25mol\) \(\Rightarrow m_{Fe}=0,25\cdot56=14\left(g\right)\)

\(\Rightarrow\%m_{Fe}=\dfrac{14}{20}\cdot100\%=70\%\) \(\Rightarrow\%m_{Ag}=30\%\)

d) Sửa đề cho dễ làm: "dd HCl 7,3%"

Theo PTHH: \(n_{HCl}=2n_{Fe}=0,5mol\) 

\(\Rightarrow m_{ddHCl}=\dfrac{0,5\cdot36,5}{7,3\%}=250\left(g\right)\) \(\Rightarrow V_{HCl}=\dfrac{250}{1,03}\approx242,72\left(ml\right)\)

Các bạn chuyên hóa giúp mình với ạ

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

\(a,Fe+2HCl\rightarrow FeCl_2+H_2\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\%\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

a) nH2=0,2(mol)

PTHH: Mg + 2 HCl -> MgCl2 + H2

0,2______0,4_______0,2_____0,2(mol)

PTHH: MgO +2 HCl -> MgCl2 + H2O

b) mMgO= 8,8 - 0,2.24=4(g)

%mMgO= (4/8,8).100= 45,455%

=>%Mg=54,545%

c) nHCl(tổng)= 2. nMg + 2. nMgO= 2. 0,2+ 0,1.2=0,6(mol)

=> mHCl= 0,6.36,5=21,9(g)

=>mddHCl=(21,9.100)/7,3=300(g)

d) mddMgCl2= mddHCl + m(hỗn hợp ban đầu) - mH2

<=>mddHCl= 300+ 8,8- 0,2.2= 308,4(g)

nMgCl2=0,3(mol) => mMgCl2= 0,3.95=28,5(g)

=>C%ddMgCl2= (28,5/308,4).100=9,241%