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PTHH
CO2 + 2NaOH ---> Na2CO3 + H2O
2NaHCO3 + BaCl2 = 2NaCl + CO2 + BaCO3 + H2O Không tác dụng được với KOH Mong ủng hộ mình
a; SO2, CO2, Na2O, P2O5
b;: Fe, Fe2O3,Al2O3, Na2O, KOH.
c;,Al2O3; HCl,SO2, CO2
Bài 2:
nNaOH = 1 . 0,3 = 0,3 mol
nH2SO4 = 0,5 . 0,2 = 0,1 mol
Pt: 2NaOH + H2SO4 --> Na2SO4 + 2H2O
....0,2 mol<---0,1 mol--> 0,1 mol
Xét tỉ lệ mol giữa NaOH và H2SO4:
\(\dfrac{0,3}{2}>\dfrac{0,1}{1}\)
Vậy NaOH dư
CM Na2SO4 = \(\dfrac{0,1}{0,3+0,2}=0,2M\)
CM NaOH = \(\dfrac{0,3-0,2}{0,3+0,2}=0,2M\)
a.
MgO + 2HCl → MgCl2 + H2O
2Na + 2HCl → 2NaCl + H2
Fe + 2HCl → FeCl2 + H2
b.
2KOH + CuSO4 → Cu(OH)2 + K2SO4
2Na + 2H2O → 2NaOH + H2
2NaOH + CuSO4 → Cu(OH)2 + Na2SO4
CuSO4 + BaCl2 → BaSO4 + CuCl2
Fe + CuSO4 → FeSO4 + Cu
c.
HCl + NaOH → NaCl + H2O
CO2 + 2NaOH → Na2CO3 + H2O
CO2 + NaOH → NaHCO3
a)MgO+2HCl--->MgCl2+H2
KOH+HCl--->KCl+H2O
2Na+2HCl--->2NaCl+H2
Fe+2HCl--->FeCl2+H2
b)2Na+2H2O--->2NaOH+H2
2NaOH+CuSO4--->Na2SO4+Cu(OH)2
BaCl2+CuSO4--->BaSO4+CuCl2
Fe+CuSO4--->FeSO4+Cu
c)HCl+NaOH--->NaCl+H2O
CO2+2NaOH--->Na2CO3+H2O
CO2+NaOH--->NaHCO3
MgSO4+2NaOH--->Na2SO4+Mg(OH)2
a) PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
b) Ta có: \(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,2\cdot95=19\left(g\right)\\C\%_{HCl}=\dfrac{0,4\cdot36,5}{200}\cdot100\%=7,3\%\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{MgO}+m_{ddHCl}=208\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{19}{208}\cdot100\%\approx9,13\%\)
c) PTHH: \(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_2\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{NaOH}=\dfrac{200\cdot4\%}{40}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) \(\Rightarrow\) NaOH p/ứ hết, MgCl2 còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,2\left(mol\right)\\n_{Mg\left(OH\right)_2}=0,1\left(mol\right)=n_{MgCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{MgCl_2\left(dư\right)}=9,5\left(g\right)\\m_{Mg\left(OH\right)_2}=0,1\cdot58=5,8\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{ddA}+m_{ddNaOH}-m_{Mg\left(OH\right)_2}=402,2\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{402,2}\cdot100\%\approx2,91\%\\C\%_{MgCl_2\left(dư\right)}=\dfrac{9,5}{402,2}\cdot100\%\approx2,36\%\end{matrix}\right.\)
Theo gt ta có: $n_{MgO}=0,2(mol)$
a, $MgO+2HCl\rightarrow MgCl_2+H_2O$
b, Ta có: $n_{HCl}=0,4(mol)\Rightarrow x=7,3$
Bảo toàn khối lượng ta có: $m_{ddA}=208(g)$
$\Rightarrow \%C_{MgCl_2}=9,13\%$
c, Ta có: $n_{NaOH}=0,2(mol)$
$\Rightarrow n_{Mg(OH)_2}=0,1(mol)$
Bảo toàn khối lượng ta có: $m_{ddB}=208+200-0,1.58=402,2(g)$
$\Rightarrow \%C_{MgCl_2}=2,36\%$
a,
\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\left(1\right)\)
\(NaOH+HCl\rightarrow NaCl+H_2O\left(2\right)\)
b,
Ta có :
\(n_{NaCl}=\frac{20,475}{58,5}=0,35\left(mol\right)\)
\(n_{Na2CO3}=\frac{10,6}{106}=0,1\left(mol\right)\)
\(\Rightarrow n_{NaCl\left(1\right)}=0,1.2=0,2\left(mol\right)\)
\(n_{NaCl}=0,35-0,2=0,15\left(mol\right)\)
\(\Rightarrow n_{NaOH}=0,15\left(mol\right)\)
\(\Rightarrow x=CM_{NaOH}=\frac{0,15}{0,3}=0,5M\)
a)SO2+H2O->H2SO3
BaO+H2O->Ba(OH)2
b)Ca(OH)2+SO2->CaSO3+H2O
Ca(OH)2+2SO2->Ca(HSO3)2
c)Fe2O3+6HCl->2FeCl3+3H2O
BaO+2HCl->BaCl2+H2O
a) \(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)
b) \(SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\)
c) \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(BaO+2HCl\rightarrow BaCl_2+H_2O\)
a) Các chất td được với dd HCl: MgO, KOH, Na, MgSO4, Fe
Pt: MgO + 2HCl --> MgCl2 + H2O
.....KOH + HCl --> KCl + H2O
....2Na + 2HCl --> 2NaCl + H2
.....MgSO4 + 2HCl --> MgCl2 + H2SO4
.....Fe + 2HCl --> FeCl2 + H2
b) Các chất td được với CuSO4: Na, BaCl2, Fe
Pt: 2Na + 2H2O --> 2NaOH + H2
......CuSO4 + 2NaOH --> Cu(OH)2 + Na2SO4
......BaCl2 + CuSO4 --> BaSO4 + CuCl2
......Fe + CuSO4 --> FeSO4 + Cu
c) Các chất td được với NaOH: HCl, CO2, MgSO4
Pt: NaOH + HCl --> NaCl + H2O
.....\(\left\{{}\begin{matrix}CO_2+NaOH\rightarrow NaHCO_3\\CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\end{matrix}\right.\)
......MgSO4 + 2NaOH --> Mg(OH)2 + Na2SO4
a)MgO+2HCl--->MgCl2+H2
KOH+HCl--->KCl+H2O
2Na+2HCl--->2NaCl+H2
Fe+2HCl--->FeCl2+H2
b)2Na+2H2O--->2NaOH+H2
2NaOH+CuSO4--->Na2SO4+Cu(OH)2
BaCl2+CuSO4--->BaSO4+CuCl2
Fe+CuSO4--->FeSO4+Cu
c)HCl+NaOH--->NaCl+H2O
CO2+2NaOH--->Na2CO3+H2O
CO2+NaOH--->NaHCO3
MgSO4+2NaOH--->Na2SO4+Mg(OH)2