Bài 24. Ôn tập học kì I

a) \(m_{CO_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

PTHH: MgCO₃ + H₂SO₄ --> MgSO₄ + CO₂ + H₂O

_____0,02<------0,02<-----------------0,02

=> m_MgCO3 = 0,02.84 = 1,68 (g)

=> \(\left\{{}\begin{matrix}\%MgCO_3=\dfrac{1,68}{10,3}.100\%=16,31\%\\\%MgCl_2=\dfrac{10,3-1,68}{10,3}.100\%=83,69\%\end{matrix}\right.\)

b) m_H2SO4 = 0,02.98 = 1,96 (g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{1,96}{100}.100\%=1,96\%\)

a) K₂CO₃ + 2HCl --> 2KCl + CO₂ + H₂O

b) \(n_{K_2CO_3}=\dfrac{13,8}{138}=0,1\left(mol\right)\)

PTHH: K₂CO₃ + 2HCl --> 2KCl + CO₂ + H₂O

______0,1----->0,2------>0,2--->0,1

=> V_CO2 = 0,1.22,4 = 2,24 (l)

c) m_HCl = 0,2.36,5 = 7,3 (g)

\(m_{ddHCl}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)

d) m_KCl = 0,2.74,5 = 14,9 (g)

m_{dd sau pư} = 13,8 + 100 - 0,1.44 = 109,4 (g)

=> \(C\%\left(KCl\right)=\dfrac{14,9}{109,4}.100\%=13,62\%\)

rất mog nhận đc sự giúp đỡ từ mn

- Cho 3 dd tác dụng với quỳ tím:

+ QT chuyển đỏ: HCl, H₂SO₄

+ QT chuyển xanh: KOH

- Cho 2 dd còn lại tác dụng với dd BaCl₂

+ Kết tủa trắng: H₂SO₄

\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

+ Không hiện tượng: HCl

1) \(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)

2) \(P_2O_5+3Ca\left(OH\right)_2\rightarrow Ca_3\left(PO_4\right)_2+3H_2O\)

3) \(Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow2AlCl_3+3BaSO_4\downarrow\)

4) \(Zn\left(OH\right)_2\underrightarrow{t^o}ZnO+H_2O\)

5) \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

6) \(Zn+FeSO_4\rightarrow ZnSO_4+Fe\)

a) CaO + 2HCl --> CaCl₂ + H₂O

CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

b) \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

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

_____0,1<-----0,2<-------0,1<-----0,1

=> m_CaCO3 = 0,1.100 = 10 (g)
=> m_CaO = 15,6 - 10 = 5,6 (g)

b) \(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH:CaO + 2HCl --> CaCl₂ + H₂O

_____0,1--->0,2------>0,1

=> m_HCl = (0,2+0,2).36,5 = 14,6 (g)

=> \(m_{ddHCl}=\dfrac{14,6.100}{14,6}=100\left(g\right)\)

m_{dd sau pư} = 15,6 + 100 - 0,1.44 = 111,2 (g)

=> \(C\%\left(CaCl_2\right)=\dfrac{\left(0,1+0,1\right).111}{111,2}.100\%=19,96\%\)

PTHH : CaO + 2HCl ---> CaCl₂ + H₂O (1)

             1       :   2       :       1       :  2

CaCO₃ + 2HCl ---> CaCl₂ + H₂O + CO₂ (2)

1           :   2        :       1       :  1       :  1

Ta có \(n_{CO_2}=\dfrac{V}{22.4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

=> \(n_{CaCO_3}=0,1\left(mol\right)\)

=> \(m_{CaCO_3}=n.M=0,1.100=10\left(g\right)\)

=> m_CaO = 15,6 - 10 = 5,6 (g) 

c) \(n_{CaO}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(m_{CO_2}=n.M=0,1.44=4,4\left(g\right)\)

Ta có \(m_{HCl}=m_{HCl\left(1\right)}+m_{HCl\left(2\right)}\)

\(=n_{HCl\left(1\right)}.M+n_{HCl\left(2\right)}.M\)

\(0,2.36,5+0,2.36,5=14,6\left(g\right)\)

=> \(m_{ddHCl}=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{14,6.100\%}{14.6\%}100\left(g\right)\)

\(m_{dd\text{ sau pư}}=m_{ddHCl}+m_{CaO}+m_{CaCO_3}-m_{CO_2}\)

= 100 + 5.6 + 10 - 4,4 = 111.2(g)

=> \(m_{CaCl_2}=m_{CaCl_2\left(1\right)}+m_{CaCl_2\left(2\right)}\)

\(=n_{CaCl_2\left(1\right)}.M+n_{CaCl_2\left(2\right)}.M\)

= 0,1.91 + 0,1.91 = 18,2 (g)

=> \(C\%=\dfrac{m_{CaCl_2}}{m_{\text{dd sau pư}}}.100\%=\dfrac{18,2}{111,2}.100\%=16,37\%\)

a) m_{dd CuSO4} = 25.1,12 = 28 (g)

=> \(m_{CuSO_4}=\dfrac{28.15}{100}=4,2\left(g\right)\)

PTHH: Fe + CuSO₄ --> FeSO₄ + Cu

_____a----->a---------->a--------->a

=> 2,5 + 64a - 56a = 2,58

=> a = 0,01 (mol)

=> m_{CuSO4 dư} = 4,2 - 0,01.160 = 2,6 (g)

b) m_{dd sau pư} = 28 + 0,01.56 - 0,01.64 = 27,92(g)

\(\left\{{}\begin{matrix}C\%\left(CuSO_4\right)=\dfrac{2,6}{27,92}.100\%=9,3\%\\C\%\left(FeSO_4\right)=\dfrac{0,01.152}{27,92}.100\%=5,44\%\end{matrix}\right.\)