Bài 45: Bài thực hành 7

PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

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

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\). ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

b, Ta có: \(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)

c, \(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5M\)

\(C_{M_{ZnSO_4}}=\dfrac{0,1}{0,2}=0,5M\)

Bạn tham khảo nhé!

a) mHCl= 200.10%=20(g)

b) mHCl=120.10%=12(g)

c) nHCl=0,1.0,5=0,05(mol)

=>mHCl=0,05.36,5=1,825(g)

d) nHCl=0,2.0,2=0,04(mol)

=> mHCl=0,04.36,5= 1,46(g)

ta co pthh Mg(NO3)2 + H2O \(\rightarrow\) MgO + 2HNO3

3 Al2(SO4)3 + 6 H2O \(\rightarrow\) 6AlO +6H2SO4

ta co nong do dd cua Al2(SO4)3= \(\dfrac{n}{v}\)\(\rightarrow\)nAl2(SO4)3= CM*V= 2.4* \(\dfrac{300}{1000}\)=0.72 mol

theo pthh nH2O= \(\dfrac{6}{3}\)nAl2(SO4)3= \(\dfrac{6}{3}\).0.72=1.44 mol

\(\Rightarrow\)VH2O= 1.44*22.4\(\approx\)32.26 l

\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ n_{Zn} = \dfrac{13}{65} = 0,2 < n_{H_2SO_4} = \dfrac{200.24,5\%}{98} = 0,5 \to H_2SO_4\ dư\\ n_{H_2SO_4\ pư} =n_{Zn} = 0,2(mol)\\ \Rightarrow m_{H_2SO_4\ dư} = (0,5 - 0,2).98 = 29,4(gam)\\ c) n_{FeSO_4} = n_{H_2} = n_{Zn} = 0,2(mol)\\ m_{FeSO_4} = 0,2.152 = 30,4(gam)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\)

a/ \(Zn\left(0,1\right)+2HCl\left(0,2\right)\rightarrow ZnCl_2\left(0,1\right)+H_2\left(0,1\right)\)

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

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

c/ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

d/ Chất còn lại trong dung dịch sau phản ứng là ZnCl₂

\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

1.Fe2O3+3CO- > 2Fe + 3CO2

2.3AgNO3+Al -> Al(NO3)3+ 3Ag

3.2HCl+CaCO3 -> CaCl2+H2O+ CO2

4.2C4H10+5O2 -> 4CO2+2H2O

5.NaOH+FE2(SO4)3 -> Fe(OH)3+Na2SO4

6.4FeS2+11O2 -> 2Fe2O3+8SO2

7.6KOH+Al2(SO4)3 -> 3K2SO4+2Al(OH)3

9.8Al+3Fe3O4 -> 4Al2O3+9Fe

10.Fe3O4+4CO -> 3Fe+4CO2

11.FeO+H2 -> Fe+H2O

11.FeO+H2 -> Fe+H2O

PTHH 1: Fe₂O₃ + 3CO -t^o-> 2Fe + 3CO₂

PTHH 2: 3AgNO₃ + Al -> Al(NO₃)₃ + 3Ag

PTHH 3: 2HCl + CaCO₃ -> H₂O + CO₂ + CaCl₂

PTHH 4: 2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O

PTHH 5: 6NaOH + Fe₂(SO₄)₃ -> 2Fe(OH)₃ + 3Na₂SO₄

PTHH 6: 4FeS₂ + 11O₂ -t^o-> 2Fe₂O₃ + 8SO₂

3CO + Fe2O3 → 2Fe +
3CO2

3AgNO3 + Al → 3Ag +
Al(NO3)3

CaCO3 + 2HCl → H2O + CO2 + CaCl2

13O2 + 2C4H10 → 10H2O +
8CO2

Fe2(SO4)3 + 6NaOH → 3Na2SO4 + 2Fe(OH)3

4FeS2 + 11O2 → 2Fe2O3 + 8SO2

Al2(SO4)3 + 6KOH → 2Al(OH)3 + 3K2SO4

12CH4 + 5H2O + 5O2 → 9CO + 29H2 + 3CO2

8Al + 3Fe3O4 → 4Al2O3 + 9Fe

FeO + H2 → Fe + H2O

\(2K\left(0,1\right)+2H_2O\left(0,1\right)\rightarrow2KOH\left(0,1\right)+H_2\)

\(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)

\(\Rightarrow m_{H_2O\left(pứ\right)}=0,1.18=1,8\left(g\right)\)

\(\Rightarrow m_{H_2O\left(dư\right)}=101,8-1,8=100\left(g\right)\)

\(\Rightarrow m_{KOH}=0,1.56=5,6\left(g\right)\)

\(\Rightarrow C\%\left(KOH\right)=\dfrac{5,6}{100+5,6}.100\%=5,3\%\)

Gọi khối lượng BaCl₂ cần thêm vào là: x (g)

Ta có: \(m_{BaCl_2\left(150\%\right)}=27.10\%=2,7\left(g\right)\)

\(\Rightarrow m_{BaCl_2}=x+2,7\left(g\right)\)

\(\Rightarrow m_{ddBaCl_2}=x+27\left(g\right)\)

\(\Rightarrow\dfrac{x+2,7}{x+27}=0,25\)

\(\Rightarrow x=5,4\left(g\right)\)

Sửa thành 2,24 gam cho số đẹp bạn nhé!

Ta có: \(n_{Fe}=\dfrac{2,24}{56}=0,04\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

___0,04__0,08____0,04__0,04 (mol)

Ta có: \(m_{HCl}=0,08.36,5=2,92\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{2,92}{14,6\%}=20\left(g\right)\)

Ta có: m _{dd sau pư} = m_Fe + m _{dd HCl} - m_H2 = 2,24 + 20 - 0,04.2 = 22,16 (g)

\(\Rightarrow C\%_{FeCl_2}=\dfrac{0,04.127}{22,16}.100\%\approx22,9\%\)

Bạn tham khảo nhé!