Câu 3.

a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:     0,1     0,1            0,1        0,1

b,\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c,\(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)

\(m_{ddsaupư}=6,5+\dfrac{0,1.98.100}{25}-0,1.2=45,5\left(g\right)\)

\(\Rightarrow C\%_{ddZnSO_4}=\dfrac{16,1.100\%}{45,5}=35,4\%\)

n Zn= 19,5/65=0,3 (mol).

PTPƯ: Zn(0.3) +   HCl(0.6) ---->  ZnCl2(0.3) + H2(0,3)

mHCl=0,6.36.5=21.9(g)

a) C%HCl= 21.9/300.100%=7,3%

b) VH2=0,3.22,4=6,72(lít)

c) mH2=0,3.2=0,6(g)

mZnCl2=0,3.136=40,8(g)

mddZnCl2 =(19,5+300)-0,6=318,9(g)

C%=mZnCl2/mddZnCl2.100= 40,8/318,9.100=12,793%

Fe+H2SO4->FeSO4+H2

0,15---0,15-----0,15---0,15 mol

n Fe=8,4\56=0,15 mol

=>VH2=0,15.22,4=3,36l

=>m H2SO4=0,15.98=14,7g

=>C% H2SO4=14,7\245 .100=6%

=>m dd muối=8,4+245-0,15.2=253,1g

=>C% muối =0,15.152\253,1 .100=9%

`a)PTHH:`

  `Zn + 2HCl -> ZnCl_2 + H_2`

`0,02`                       `0,02`    `0,02`        `(mol)`

`n_[Zn]=[1,3]/65=0,02(mol)`

`b)V_[H_2]=0,02.22,4=0,448(l)`

`c)C%_[ZnCl_2]=[0,02.136]/[1,3+50-0,02.2].100~~5,31%`

\(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)

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

0,02   0,04         0,02       0,02  ( mol )

\(V_{H_2}=0,02.22,4=0,448\left(l\right)\)

\(C\%_{ZnCl_2}=\dfrac{0,02.136}{1,3+50-0,02.2}.100=5,3\%\)

\(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
         0,02                 0,02           0,02 
\(V_{H_2}=0,02.22,4=0,448\left(l\right)\\ m_{\text{dd}}=1,3+50-\left(0,02.2\right)=51,26\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,02.136}{51,26}.100\%=5,3\%\)

Bài 4 : 

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

            1          2              1          1

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

b) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)⁰/₀

d) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)

\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)⁰/₀

 Chúc bạn học tốt

Bài 3 : 

\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)

a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)

            1            1                1           1

          0,5          0,5            0,5          0,5

b) \(n_{H2}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)

c) \(n_{H2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

⇒ \(m_{H2SO4}=0,5.98=49\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{49.100}{200}=24,5\)⁰/₀

d) \(n_{MgSO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

⇒ \(m_{MgSO4}=0,5.120=60\left(g\right)\)

\(m_{ddspu}=12+200-\left(0,5.2\right)=211\left(g\right)\)

\(C_{MgSO4}=\dfrac{60.100}{211}=28,44\)⁰/₀

 Chúc bạn học tốt

Mg+H2SO4->MgSO4+H2

0,5----0,5--------0,5--------0,5

n Mg=12\24=0,5 mol

=>VH2=0,5.22,4=11,2l

=>m H2SO4=0,5.98=49 g

=>C% =49\200.100=24,5g

=>m dd=12+200-0,5.2=211g

=>C%=0,4.120\211.100=22,74885%

a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

   \(m_{H_2SO_4}=147.10\%=14,7\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:     0,1       0,1            0,1       0,1

Ta có: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\) ⇒ Zn hết, H₂SO₄ dư

b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c) m_{dd sau pứ} = 6,5 + 147 - 0,1.2 = 153,3 (g)

\(C\%_{ddZnSO_4}=\dfrac{0,1.161.100\%}{153,3}=10,502\%\)

\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,15-0,1\right).98.100\%}{153,3}=3,196\%\) 

Câu 1:

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

\(n_{H_2}=n_{Zn}=0,25\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,25.24,79=6,1975\left(l\right)\)

c, \(n_{HCl}=2n_{Zn}=0,5\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{18,25}{10\%}=182,5\left(g\right)\)

d, \(n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,25.136}{16,25+182,5-0,25.2}.100\%\approx17,15\%\)

Câu 2:

a, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

c, \(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)

\(\Rightarrow V_{H_2SO_4}=\dfrac{0,5}{2}=0,25\left(l\right)\)

d, \(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,5}{0,25}=2\left(M\right)\)

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)

b, \(n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{0,3.98}{120}.100\%=24,5\%\)

c, m _{dd sau pư} = 16,8 + 120 - 0,3.2 = 136,2 (g)

d, \(n_{FeSO_4}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,3.152}{136,2}.100\%\approx33,48\%\)

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