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\(n_{HCl}=0,2.2=0,04\left(mol\right)\)
Pt: \(NaOH+HCl\rightarrow NaCl+H_2O\)
0,04mol \(\leftarrow\) 0,04mol \(\rightarrow\) 0,04mol
\(V_{NaOH}=\dfrac{0,04}{0,1}=0,4\left(l\right)\)
\(\Sigma_{V\left(spu\right)}=0,2+0,4=0,6\left(l\right)\)
\(C_{M_{NaCl}}=\dfrac{0,04}{0,6}=0,067M\)
b) Pt: \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
0,02mol \(\leftarrow\) 0,04mol\(\rightarrow\) 0,02mol
\(m_{Ca\left(OH\right)_2}=0,02.74=1,48\left(g\right)\)
\(m_{dd}=\dfrac{1,48.100}{5}=29,6\left(g\right)\)
\(C\%_{CaCl_2}=\dfrac{0,02.111.100}{29,6}=7,5\%\)
nHCl=CM.V=1.0,6=0,6(mol)
Pt: HCl+ NaOH-> NaCl + H2O
cứ:1.............1.............1 (mol)
vậy:0,6----->0,6------>0,6(mol)
=> mNaOH=0,6.40=24(g)
\(\Rightarrow m_{ddNaOH}=\dfrac{m_{NaOH}.100\%}{C\%}=\dfrac{24.100}{30}=80\left(g\right)\)
b) Nếu thay NaOH bằng Ca(OH)2 thì ta có PT:
PT: 2HCl + Ca(OH)2 -> CaCl2 +2H2O
Cứ:2................1.................1 (mol)
Vậy: 0,6------->0,3----------->0,3(mol)
=> mCa(OH)2=n.M=0,3.74=22,2(g)
\(\Rightarrow m_{ddCa\left(OH\right)_2}=\dfrac{m_{Ca\left(OH\right)_2}.100\%}{C\%}=\dfrac{22,2.100}{7,351}\approx302\left(g\right)\)
\(\Rightarrow V_{ddCa\left(OH\right)_2}=\dfrac{m_{ddCa\left(OH\right)_2}}{D}=\dfrac{302}{1,045}\approx289\left(ml\right)=0,289\left(lít\right)\)
Ta có : \(n_{OH}=2C_{MBa\left(OH\right)2}V+C_{MKOH}V=2,09\left(mol\right)\)
\(BTNT\left(O\right):n_{H2O}=n_{OH}=2,09\left(mol\right)\)
\(BTNT\left(H\right):n_H=n_{H2O}=2,09\left(mol\right)\)
Mà \(n_H=2,09=1,98V+2,2V=4,18V\)
\(\Rightarrow V=0,5\left(l\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Ba\left(OH\right)2}=0,76\\n_{H2SO4}=0,55\end{matrix}\right.\) ( mol )
\(PTHH:H_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4+2H_2O\)
Thấy ; \(0,55< 0,76\)
\(\Rightarrow n_{BaSO4}=0,55\left(mol\right)\)
\(\Rightarrow m_{kt}=128,15g\)
\(n_{Ca\left(OH\right)_2}=0,3.1=0,3\left(mol\right)\\ n_{HCl}=0,2.0,2=0,04\left(mol\right)\)
a
\(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
0,02<------0,04----->0,02
Xét \(\dfrac{0,3}{1}>\dfrac{0,04}{2}\Rightarrow Ca\left(OH\right)_2.dư\)
\(m_{CaCl_2}=0,02.111=2,22\left(g\right)\)
b
Muốn pứ xảy ra hoàn toàn phải thêm dung dịch HCl 0,2 M
\(n_{HCl.cần}=2n_{Ca\left(OH\right)_2}=0,3.2=0,6\left(mol\right)\\ n_{HCl.cần.thêm}=0,6-0,04=0,56\left(mol\right)\)
\(V_{cần.\left(HCl\right)}=\dfrac{0,56}{0,2}=2,8\left(l\right)=280\left(ml\right)\\ V_{cần.thêm\left(HCl\right)}=280-200=80\left(ml\right)\)
c
\(CM_{CaCl_2}=\dfrac{0,02}{0,3+0,28}=\dfrac{1}{29}M\)
\(n_{NaOH}=0,25.V\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=0,5.V\left(mol\right)\)
=> \(n_{OH^-}=0,25.V+2.0,5.V=1,25V\left(mol\right)\)
\(n_{HCl}=0,55.2=1,1\left(mol\right)=>n_{H^+}=1,1\left(mol\right)\)
H+ + OH- --> H2O
1,1->1,1
=> 1,25.V = 1,1
=> V = 0,88(l)
2NaOH + MgCl2 => Mg(OH)2 ↓ + 2NaCl
2KOH + MgCl2 => 2KCl + Mg(OH)2 ↓
Ca(OH)2 + MgCl2 => CaCl2 + Mg(OH)2 ↓
Mg(OH)2 + 2HCl => MgCl2 + 2H2O
NaOH + HCl => NaCl + H2O
KOH + HCl => KCl + H2O
Ca(OH)2 + 2HCl => CaCl2 + 2H2O
Cu(OH)2 + 2HCl => CuCl2 + 2H2O
Na2CO3 + 2HCl => 2NaCl + H2O + CO2 ↑
2NaOH + H2SO4 => Na2SO4 + 2H2O
2KOH + H2SO4 => K2SO4 + 2H2O
Ca(OH)2 + H2SO4 => CaSO4 + 2H2O
Mg(OH)2 + H2SO4 => MgSO4 + 2H2O
Cu(OH)2 + H2SO4 => CuSO4 + 2H2O
Na2CO3 + H2SO4 => Na2SO4 + H2O + CO2 ↑
Na2CO3 + BaCl2 => BaCO3 ↓ + 2NaCl
MgCl2 + Na2CO3 => MgCO3 ↓ + 2NaCl
BaCl2 + H2SO4 => BaSO4 ↓ + 2HCl
Ca(OH)2 + Na2CO3 => 2NaOH + CaCO3 ↓