Bài 41: Độ tan của một chất trong nước

giúp e với mn ơi

20 độ C:

Khối lượng chất tan muối K₂SO₄ là:

\(m_{K_2SO_4}=\dfrac{666,6.11,1}{111,1}=66,6\left(g\right)\)

a) $n_{Zn} = \dfrac{26}{65} = 0,4(mol)$

$Zn + 2HCl \to ZnCl_2 + H_2$

Theo PTHH : $n_{HCl} = 2n_{Zn} = 0,8(mol) \Rightarrow V_{dd\ HCl} = \dfrac{0,8}{1} = 0,8(lít)$

b)

$ZnCl_2 + 2AgNO_3 \to 2AgCl + Zn(NO_3)_2$

$n_{AgNO_3} = 2n_{ZnCl_2} = 0,8(mol)$

$V_{dd\ AgNO_3} = \dfrac{0,8}{0,5} = 1,6(lít)$

c)

$n_{AgCl} = n_{AgNO_3} = 0,8(mol) \Rightarrow m_{AgCl} = 0,8.143,5 = 114,8(gam)$

\(m_{H_2O}=37,6.1=37,6\left(g\right)\\ m_{ddNaOH}=12,4+37,6=50\left(g\right)\\ C\%_{ddNaOH}=\dfrac{12,4}{50}.100=24,8\%\)

Ta có: \(D_{H_2O}=1\left(g/mol\right)\Rightarrow m_{H_2O}=37,6.1=37,6\left(g\right)\)

\(\Rightarrow C\%_{ddNaOH}=\dfrac{12,4.100\%}{37,6}=32,98\%\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\\C_{M_{FeCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)

\(SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\)

\(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)

Theo pt: \(nCaSO_3=nSO_2=0,3mol\)

=> \(mCaSO_3=36g\)

Câu 2 : 

\(n_{Zn}=\dfrac{3.25}{65}=0.05\left(mol\right)\)

\(n_{HCl}=0.12\cdot1=0.12\left(mol\right)\)

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

\(1...........2\)

\(0.05.......0.15\)

\(LTL:\dfrac{0.05}{1}< \dfrac{0.12}{2}\Rightarrow HCldư\)

\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)

Câu 3 : 

\(n_{HCl}=0.2\cdot3=0.6\left(mol\right)\)

\(M+2HCl\rightarrow MCl_2+H_2\)

\(0.3..........0.6\)

\(M_M=\dfrac{7.2}{0.3}=24\left(g\text{/}mol\right)\)

\(M:Mg\left(Magie\right)\)

Câu 4 : 

\(n_{FeO}=n_{Fe_2O_3}=n_{Fe_3O_4}=n_{Cu_2O}=1\left(mol\right)\)

\(FeO+H_2\underrightarrow{^{^{t^0}}}Fe+H_2O\)

\(1..........1..........1\)

1 (mol) H₂ => 1 (mol) Fe (1) 

\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)

\(1..............3...........2\)

3 (mol) H₂ => 2 (mol) Fe 

=> 1 (mol) H₂ => \(\dfrac{2}{3}\) (mol) Fe (2) 

\(Fe_3O_4+4H_2\underrightarrow{^{^{t^0}}}3Fe+4H_2O\)

\(1..............4.........3\)

4 (mol) H₂ => 3 (mol) Fe 

=> 1 (mol) H₂ => 0.75 (mol) Fe (3)

\(Cu_2O+H_2\underrightarrow{^{^{t^0}}}2Cu+H_2O\)

\(1............1.........2\)

1 (mol) H₂ => 2 (mol) Cu (4)

Từ (1) , (2) , (3) , (4) : 

Nếu lấy cùng lượng số mol với oxit Cu₂O thì tỉ lệ số mol H₂ và kim loại là lớn nhất.

1.B

2.D

3.C

4.A

5.C

Câu 1: B

Câu 2: D

-Chúc bạn học tốt-

\(a.\)

\(m_{MgCl_2}=25\cdot8\%=2\left(g\right)\)

\(b.\)

\(n_{K_2SO_4}=0.25\cdot0.2=0.05\left(mol\right)\)

\(m_{K_2SO_4}=0.05\cdot174=8.7\left(g\right)\)

\(n_{CaO}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(CaO+2HCl\rightarrow CaCl_2+H_2O\)

\(0.2...........0.4.........0.2\)

\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)

\(C\%_{HCl}=\dfrac{14.6}{200}\cdot100\%=7.3\%\)

\(m_{\text{dung dịch sau phản ứng}}=11.2+200=211.2\left(g\right)\)

\(m_{CaCl_2}=0.2\cdot111=22.2\left(g\right)\)

\(C\%_{CaCl_2}=\dfrac{22.2}{211.2}\cdot100\%=10.51\%\)