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

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

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

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

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

a) 2Al + 6HCl -> 2AlCl3 + 3H2

Al2O3 + 6HCl -> 2AlCl3 + 3H2O

nH2 = 0,15mol => nAl=0,1mol => mAl=2,7g; mAl2O3 = 10,2g => nAl2O3 = 0,1mol

=>%mAl=20,93% =>%mAl2O3 = 79,07%

b) nHCl = 0,1.3+0,1.6=0,9 mol=>mHCl(dd)=100g

mddY=12,9+100-0,15.2=112,6g

mAlCl3=22,5g=>C%=19,98%

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

Ta có: \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{H_2}=0,03\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,03.24=0,72\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,72}{1,74}.100\%\approx41,38\%\\\%m_{AlCl_3}\approx58,62\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,06.36,5=2,19\left(g\right)\)

\(\Rightarrow C\%_{ddHCl}=\dfrac{2,19}{500}.100\%=0,438\%\)

Bạn tham khảo nhé!

a, nH2 = 0,03 ( mol )

=> nMg = nH2 = 0,03 ( mol )

=> mMg = 0,72 g

=> %Mg \(\approx\) 41,38 % .

=> % Al \(\approx\) 58,62 % .

b, Có : nH2 = 0,03 mol

=> nHCl = nHCl_{từ Al2O3} + nHCl_{từ Mg} = 0,06 + 0,06 = 0,12 ( mol )

=> mHCl = 4,38 ( g )

Lại có : mdd = mhh + mddHCl = 501,74 ( g )

=> \(C\%=\dfrac{m_{HCl}}{m_{dd}}.100\%\approx0,87\%\)

( chắc đoạn trên là Al2O3 :vvvv )

chj kb vs em nha em ib rồi ạ

1.

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

x_____2x_____________x

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

y_____3y_____________1,5y

\(n_{H2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)

Gọi x, y lần lượt là nMg và nAl, ta có:

\(\left\{{}\begin{matrix}24x+27y=10,2\\x+1,5y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)

Bạn tự tính % nhé.

\(n_{HCl}=2x+3y=1\left(mol\right)\)

\(\Rightarrow CM_{HCl}=\frac{1}{0,5}=2M\)

2.

Đổi 500 ml = 0,5l

\(n_{NaOH}=0,5.4=2\left(mol\right)\)

\(n_{MnO2}=\frac{69,6}{87}=0,8\left(mol\right)\)

\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\left(1\right)\)

\(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\left(2\right)\)

Theo (1) nCl2 = nMnO2 = 0,8(mol)

Lập tỉ lệ \(\Rightarrow\) NaOH dư

Theo (2) nNaOH(pư) = 2nCl2 = 1,6(mol)

\(\Leftrightarrow n_{NaOH_{dư}}=2-1,6=0,4\left(mol\right)\)

Theo (2) nNaCl = nCl2 = 0,8(mol)

\(\Rightarrow CM_{NaCl}=\frac{0,8}{0,5}=1,6M\)

\(CM_{NaOH.dư}=\frac{0,4}{0,5}=0,8M\)

3.\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

\(m_{dd_{HCl}}=100,84.1,19=120\left(g\right)\)

\(n_{HCl}=\frac{120.36,5}{36,5}=1,2\left(mol\right)\)

\(n_{H2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)

\(\Rightarrow n_{HCl\left(1\right)}=0,4.2=0,8\left(mol\right);n_{HCl\left(2\right)}=1,2-0,8=0,4\left(mol\right)\)

\(n_{ZnO}=0,2\left(mol\right)\)

\(m_{Zn}=0,4.65=26\left(g\right)\)

\(m_{ZnO}=0,2.81=16,2\left(g\right)\)