\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100\cdot20\%}{98}=\dfrac{10}{49}\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(TC:\dfrac{0.02}{1}< \dfrac{10}{49}\Rightarrow H_2SO_4dư\)

\(m_{dd}=1.6+100=101.6\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{101.6}\cdot100\%=3.15\%\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(\dfrac{10}{49}-0.02\right)\cdot98}{101.6}\cdot100\%=17.7\%\)

\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(m_{ct}=\dfrac{3,65.200}{100}=7,3\left(g\right)\)

\(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)

             1             2            1           1

           0,05        0,2         0,05

b) Lập tỉ số so sánh : \(\dfrac{0,05}{1}< \dfrac{0,2}{2}\)

                       ⇒ CuO phản ứng hết , HCl dư

                       ⇒ Tính toán dựa vào số mol của CuO

\(n_{CuCl2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

⇒ \(m_{CuCl2}=0,05.135=6,75\left(g\right)\)

\(n_{HCl\left(dư\right)}=0,2-\left(0,05.2\right)=0,1\left(mol\right)\)

⇒ \(m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

\(m_{ddspu}=4+200=204\left(g\right)\)

\(C_{CuCl2}=\dfrac{6,75.100}{204}=3,31\)⁰/₀

\(C_{HCl\left(dư\right)}=\dfrac{3,65.100}{204}=1,8\)⁰/₀

 Chúc bạn học tốt

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ a,Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=2.0,2=0,4\left(mol\right)\\ b,m_{ddHCl}=\dfrac{0,4.36,5.100}{20}=73\left(g\right)\\ c,n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\ m_{ddsau}=11,2+73-0,2.2=83,8\left(g\right)\\ C\%_{ddFeCl_2}=\dfrac{0,2.127}{83,8}.100\approx30,31\%\)

\(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\\ m_{HCl}=\dfrac{100.18,25}{100}=18,25\left(g\right)\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)  
            0,125                            0,125 (mol ) 
\(\Rightarrow V_{H_2}=0,125.22,4=2,8\left(l\right)\\ \)   
\(C\%=\dfrac{8,125}{8,125+18,25}.100\%=30,8\%\)
 

Bài 18:

Ta có: \(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\)

\(m_{HCl}=100.18,25\%=18,25\left(g\right)\Rightarrow n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Xét tỉ lệ: \(\dfrac{0,125}{1}< \dfrac{0,5}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,125\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,125.22,4=2,8\left(g\right)\)

\(m_{H_2}=0,125.2=0,25\left(g\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,125\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Zn}=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,25\left(mol\right)\)

Có: m _{dd sau pư} = 8,125 + 100 - 0,25 = 107,875 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,125.136}{107,875}.100\%\approx15,76\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,25.36,5}{107,875}.100\%\approx8,46\%\end{matrix}\right.\)

Bạn tham khảo nhé!

`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2↑`

`0,125`  `0,25`    `0,125`    `0,125`      `(mol)`

`b)n_[Zn] = [ 8,125 ] / 65 = 0,125 (mol)`

`n_[HCl] = [ [ 18,25 ] / 100 . 100 ] / [ 36,5 ] = 0,5 (mol)`

Ta có: `[ 0,125 ] / 1 < [ 0,5 ] / 2`

    `-> Zn` hết, `HCl` dư

 `=> V_[H_2] = 0,125 . 22,4 = 2,8 (l)`

 `=> m_[H_2] = 0,125 . 2 = 0,25 (g)`

`c)m_\text{dd sau p/ứ} = 8,125 + 100 - 0,25 = 107,875 (g)`

`=> C%_[ZnCl_2] = [ 0,125 . 136 ] / [ 107,875 ] . 100 ~~ 15,76%`

`=> C%_[HCl(dư)] = [ ( 0,5 - 0,25 ) . 36,5 ] / [ 107,875 ] . 100 ~~ 8,46%`

a) PTHH: CuO + H2SO4 ---> CuSO4 + H2O 

b) 
n Cu = 1,6 / 80 = 0,02 mol 

m H2SO4 = 20 . 100 / 100 = 20 g 

=> n H2SO4 = 20 / 98 = 0,204 mol 

TPT: 

1 mol : 1 mol 

0,02 mol : 0,204 mol 

=> Tỉ lệ: 0,02/1 < 0,204/1 

=> H2SO4 dư, tính toán theo CuO 

m dd sau p/ư = m dd H2SO4 + m CuO = 100 + 1,6 = 101,6 g 

TPT: n CuSO4 = n CuO = 0,02 mol 

=> m CuSO4 = 0,02 . 160 = 3,2 g 

=> C% CuSO4 = 3,2 / 101,6 . 100% = 3,15% 

n H2SO4 dư = 0,204 - 0,02 = 0,182 mol 

=> m H2SO4 dư = 0,182 . 98 =17,836 g 

=> C% H2SO4 = 17,836 / 101,6 . 100% = 17,83%

Bài dưới bạn :D

CuO+H2SO4--->CuSO4+H₂O

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%

\(n_{Fe_2O_3}=\dfrac{10}{160}=0,0625\left(mol\right)\\ n_{HCl}=\dfrac{200.18,25\%}{36,5}=1\left(mol\right)\\ Fe_2O_3+6HCl\xrightarrow[]{}2FeCl_3+3H_2O\\ TC:\dfrac{0,0625}{1}< \dfrac{1}{6}\Rightarrow HCl.dư\\ n_{FeCl_3}=0,0625.2=0,125\left(mol\right)\\ n_{HCl}=0,0625.6=0,375\left(mol\right)\\ m_{dd}=10+200=210\left(g\right)\\ C_{\%FeCl_3}=\dfrac{0,125.162,5}{210}\cdot100\approx9,67\%\\ C_{\%HCl\left(dư\right)}=\dfrac{\left(1-0,375\right).36,5}{210}\cdot100\approx10,86\%\)