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a) Các chất tác dụng được với: H2SO4, SO3, MgCl2, Cu(NO3)2, CO2
Pt: 2KOH + H2SO4 --> K2SO4 + 2H2O
......SO3 + 2KOH + H2O --> K2SO4 + 2H2O
......MgCl2 + 2KOH --> 2KCl + Mg(OH)2
......Cu(NO3)2 + 2KOH --> 2KNO3 + Cu(OH)2
.......CO2 + 2KOH --> K2CO3 + H2O
hoặc: CO2 + KOH --> KHCO3
b) Các chất tác dụng được với Ca(OH)2: H2SO4, SO3, MgCl2, Cu(NO3)2, CO2
Pt: Ca(OH)2 + H2SO4 --> CaSO4 + 2H2O
.....SO3 + Ca(OH)2 + H2O --> CaSO4 + 2H2O
......MgCl2 + Ca(OH)2 --> CaCl2 + Mg(OH)2
......Cu(NO3)2 + Ca(OH)2 --> Ca(NO3)2 + Cu(OH)2
......CO2 + Ca(OH)2 --> CaCO3 + H2O
hoặc: CO2 + Ca(OH)2 --> Ca(HCO3)2
Câu 1:
\(m_{Na_2CO_3}=\dfrac{C\%\cdot m_{d^2}}{100}=\dfrac{16,96\cdot100}{100}=16,96\left(g\right)\\ m_{BaCl_2}=\dfrac{C\%\cdot m_{d^2}}{100}=\dfrac{10,4\cdot200}{100}=20,8\left(g\right)\\ \Rightarrow n_{Na_2CO_3}=\dfrac{m}{M}=\dfrac{16,96}{106}=0,16\left(mol\right)\\ n_{BaCl_2}=\dfrac{m}{M}=\dfrac{20,8}{208}=0,1\left(mol\right)\)
\(m_{BaCO_3}=n\cdot M=0,1\cdot197=19,7\left(g\right)\\ \Rightarrow m_{d^2\text{ }sau\text{ }pứ}=\left(m_{d^2\text{ }Na_2CO_3}+m_{d^2\text{ }BaCl_2}\right)-m_{BaCO_3}\\ =\left(100+200\right)-19,7=280,3\left(g\right)\)
\(m_{Na_2CO_3\left(dư\right)}=n\cdot M=0,06\cdot106=6,36\left(g\right)\\ m_{NaCl}=n\cdot M=0,2\cdot58,5=11,7\left(g\right)\)
\(\Rightarrow C\%\left(Na_2CO_3\left(dư\right)\right)=\dfrac{m_{ct}}{m_{d^2}}\cdot100=\dfrac{6,36}{280,3}\cdot100=2,27\%\\ C\%\left(NaCl\right)=\dfrac{m_{ct}}{m_{d^2}}\cdot100=\dfrac{11,7}{280,3}\cdot100=4,17\%\)
Câu 2:
\(m_{HCl}=\dfrac{C\%\cdot m_{d^2}}{100}=\dfrac{150\cdot2,65}{100}=3,975\left(g\right)\\ \Rightarrow n_{HCl}=\dfrac{m}{M}=\dfrac{3,975}{36,5}=0,11\left(mol\right)\\ \Rightarrow C_{M\left(HCl\right)}=\dfrac{n}{V}=\dfrac{0,11}{2}=0,054\left(M\right)\)
Câu 3:
\(n_{NaOH}=C_M\cdot V=2\cdot1=2\left(mol\right)\\ \Rightarrow V_{d^2\text{ }NaOH}=\dfrac{n}{C_M}=\dfrac{2}{0,1}=20\left(l\right)\\ \Rightarrow V_{H_2O}=20-2=18\left(l\right)\)
nNaOH = 0,5 mol
2NaOH + MgCl2 → Mg(OH)2 + 2NaCl
0,5........0,25..................0,25...........0,5
CM MgCl2 = \(\dfrac{0,25}{0,3}\)= \(\dfrac{5}{6}\) (M)
Mg(OH)2 ---to---> MgO + H2O
0,25......................0,25
⇒ mMgO = 0,25.40 = 10 (g)
nNaOH=20/40=0,5(mol)
2NaOH+MgCl2--->Mg(OH)2+2NaCl
0,5________0,25___0,25
CM MgCl2=0,25/0,3=0,83M
Mg(OH)2--t*-->MgO+H2O
0,25__________0,25
=>mMgO=0,25.40=10(g)
n\(_{Fe}\) = \(\dfrac{11,2}{56}\)= 0,2 (mol)
a.
PTHH: Fe + H2SO4 ----> FeSO4 + H2\(\uparrow\)
mol: --0,2--->0,2----------->0,2---->0,2
b.V\(_{H_2}\)= 0,2 . 22,4 = 4,48 (lít)
c. m\(_{H_2SO_4}\)= 0,2 . 98 = 19,6 (g)
m\(_{ddH_2SO_4}\)= \(\dfrac{19,6.100}{4,9}\)= 400 (g)
d.m\(_{FeSO_4}\)= 0,2 . 152 = 30,4 (g)
m\(_{ddFeSO_4}\) = 11,2 + 400 - 0,2.2 = 410,8 (g)
C% FeSO\(_4\) = \(\dfrac{30,4}{410,8}\).100% = 7,4%
\(m_{MgCl_2}=150\times9,5\%=14,25\left(g\right)\)
\(\Rightarrow n_{MgCl_2}=\dfrac{14,25}{95}=0,15\left(mol\right)\)
\(m_{Na_2CO_3}=250\times5,3\%=13,25\left(g\right)\)
\(\Rightarrow n_{Na_2CO_3}=\dfrac{13,25}{106}=0,125\left(mol\right)\)
a) Hiện tượng: sau phản ứng có kết tủa trắng
PTHH: MgCl2 + Na2CO3 → 2NaCl + MgCO3↓
Ban đầu: 0,15...........0,125............................................(mol)
Phản ứng: 0,125..........0,125...........................................(mol)
Sau phản ứng: 0,025..........0..........→....0,25.......0,125......(mol)
b) \(m_{MgCO_3}=0,125\times84=10,5\left(g\right)\)
c) \(m_{dd}saupứ=150+250-10,5=389,5\left(g\right)\)
DD sau phản ứng gồm: MgCl2 dư và NaCl
\(m_{MgCl_2}dư=0,025\times95=2,375\left(g\right)\)
\(m_{NaCl}=0,25\times58,5=14,625\left(g\right)\)
\(\Rightarrow m_{ctan}saupư=2,375+14,625=17\left(g\right)\)
\(\Rightarrow C\%_{ct}saupư=\dfrac{17}{389,5}\times100\%=4,36\%\)
Thank kiu bn nhìu nha!!!