\(PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\)

+ \(\dfrac{n_{CuO\left(bđ\right)}}{n_{CuO\left(PTHH\right)}}=\dfrac{0,25}{1}=0,25\)

+ \(\dfrac{n_{HCl\left(bđ\right)}}{n_{HCl\left(PTHH\right)}}=\dfrac{0,5}{2}=0,25\)

=> pư trên vừa đủ

=> \(m_{CuCl_2}=0,25.135=33,75g\)

\(n_{Ca}=\dfrac{8}{40}=0,2mol\\ 2Ca+O_2\xrightarrow[]{t^0}2CaO\\ n_{CaO}=n_{Ca}=0,2mol\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5mol\\ CaO+2HCl\rightarrow CaCl_2+H_2O\\ \Rightarrow\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ n_{CaCl_2}=n_{CaO}=0,2mol\\ n_{HCl}=2n_{CaO}=0,4mol\\ m_{CaCl_2}=0,2.111=22,2g\\ m_{HCl.dư}=\left(0,5-0,4\right).36,5=3,65g\)

\(n_{CuO}=\frac{20}{80}=0,25\left(mol\right)\)

\(n_{HCl}=\frac{36,5}{36,5}=0,1\left(mol\right)\)

\(PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\)

Ban đầu: 0,25_____0,1

Phản ứng: 0,05____0,1_____0,05___________(mol)

Dư:_____0,2

Lập tỉ lệ: \(\frac{0,25}{1}>\frac{0,1}{2}\left(0,25>0,05\right)\)

\(m_{CuCl_2}=0,05.135=6,75\left(g\right)\)

\(m_{H_2O}=0,05.18=0,9\left(g\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{109,5.20\%}{36,5}=0,6\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,2}{1}< \dfrac{0,6}{2}\Rightarrow HCldư\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ n_{HCl\left(dư\right)}=0,6-0,2.2=0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,m_{ZnCl_2}=136.0,2=27,2\left(g\right)\\ c,m_{ddsau}=13+109,5-0,2.2=122,1\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{27,2}{122,1}.100\approx22,277\%\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,2.36,5}{122,1}.100\approx5,979\%\)

Zn + 2HCl -> ZnCl₂ + H₂ 

a, n_Zn = 13/65= 0,2(mol)

m_HCl= 109,5.20%/100%=21.9(g)

n_HCl=21,9/36,5=0,6(mol)

Theo PT n_HCl = 2n_Zn= 2.0,2= 0,4(mol)<0,6(mol)

=> HCl pư dư, Zn pư hết

Theo PT: n_H2= n_Zn =0,2(mol)

V_H2=0,2.22,4=4,48(l)

b, Theo PT: n_ZnCl2=n_Zn=0,2(mol)

m_ZnCl2= 0,2.136=27,2(g)

c, mdd sau pư= 13+109,5-0,2.2=122,1(g)

C%dd ZnCl2=27,2.100%/122,1=22,28%

n_HCl dư= 0,6-0,4=0,2(mol)

mHcl

\(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\\ m_{HCl}=\dfrac{100.18,25}{100}=18,25\left(g\right)\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)  
            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%`

\(CuO + 2HCl \rightarrow CuCl_2 + H_2O\)

\(n_{CuO} = \dfrac{8}{80}= 0,1 mol\)

\(m_{HCl}= 100 . 18,25\)%= 18,25g

\(n_{HCl}= \dfrac{18,25}{36,5}=0,5 mol\)

                    \(CuO + 2HCl \rightarrow CuCl_2 + H_2O\)

Trước pư:     0,1        0,5 

PƯ                0,1        0,2          0,1

Sau pư            0          0,3         0,1

Dung dịch sau pư: CuCl₂ và HCl dư

\(m_{dd sau pư} = m_{CuO} + m_{dd HCl}=8 +100= 108g\)

C%HCl\(=\dfrac{0,3 . 36,5}{108} . 100\)%= 10,139%

C%CuCl₂\(=\dfrac{0,1. 135}{108}.100\)%= 12,5%

\(a.n_{Zn}=\dfrac{13}{65}=0,2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,2         0,4          0,2         0,2

\(V_{H_2}=0,2.22,4=4,48l\\ b.m_{ddHCl}=\dfrac{0,4.36,5}{20}\cdot100=73g\\ c.CuO+H_2\underrightarrow{t^0}Cu+H_2O\\ n_{Cu}=n_{H_2}=0,2mol\\ m_{Cu}=0,2.64=12,8g\)