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\(\sum\)nHCl=0,2.2+0,3.4=1,6(mol)
CM dd HCl=\(\dfrac{1,6}{0,5}=3,2M\)
\(n_{HCl}=0,3.1=0,3\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{NaCl}=n_{HCl}=0,3\left(mol\right)\\ V_{\text{dd}NaOH}=V=\dfrac{0,3}{2}=0,15\left(l\right)\\ C_{M\text{dd}A}=C_{M\text{dd}NaCl}=\dfrac{0,3}{0,15+0,3}=\dfrac{2}{3}\left(M\right)\)
Gọi \(\left\{{}\begin{matrix}C_{M\left(A\right)}=aM\\C_{M\left(B\right)}=bM\end{matrix}\right.\)
Giả sử trộn 50ml dd A với 50ml dd B để thu được 100ml dd C
=> \(\left\{{}\begin{matrix}n_{NaOH}=0,05a\left(mol\right)\\n_{Ba\left(OH\right)_2}=0,05b\left(mol\right)\end{matrix}\right.\)
\(n_{BaSO_4}=\dfrac{9,32}{233}=0,04\left(mol\right)\)
nH2SO4 = 0,035.2 = 0,07 (mol)
PTHH: Ba(OH)2 + H2SO4 --> BaSO4 + 2H2O
0,04<----0,04<-------0,04
2NaOH + H2SO4 --> Na2SO4 + 2H2O
0,06<----0,03
=> \(\left\{{}\begin{matrix}0,05a=0,06\\0,05b=0,04\end{matrix}\right.\)
=> a = 1,2; b = 0,8
20 ml dd A chứa nNaOH = 0,02.1,2 = 0,024 (mol)
\(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)
PTHH: 2NaOH + Al2O3 --> 2NaAlO2 + H2O
0,024-->0,012
Ba(OH)2 + Al2O3 --> Ba(AlO2)2 + H2O
0,188<---0,188
=> \(V_{dd.B}=\dfrac{0,188}{0,8}=0,235\left(l\right)=235\left(ml\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
_____0,2---->0,2------>0,2----->0,2
VH2 = 0,2.22,4 = 4,48(l)
\(C_{M\left(ZnSO_4\right)}=\dfrac{0,2}{0,3}=0,667M\)
1)
$n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,2(mol)$
$m_{dd} = 6,2 + 193,8 = 200(gam) \Rightarrow C\%_{NaOH} = \dfrac{0,2.40}{200}.100\% = 4\%$
2)
$n_{K_2O} = \dfrac{23,5}{94} = 0,25(mol)$
$K_2O + H_2O \to 2KOH$
$n_{KOH} = 2n_{K_2O} = 0,5(mol) \Rightarrow C_{M_{KOH}} = \dfrac{0,5}{0,5} = 1M$
3) $n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,4(mol)$
$C_{M_{NaOH}} = \dfrac{0,4}{0,5} =0,8M$
4)
$Na_2SO_3 + 2HCl \to 2NaCl +S O_2 + H_2O$
Theo PTHH :
$n_{SO_2} = n_{Na_2SO_3} = \dfrac{12,6}{126} = 0,1(mol)$
$V_{SO_2} = 0,1.22,4 = 2,24(lít)$
5) $n_{CaO} = \dfrac{5,6}{56} = 0,1(mol)$
$CaO + 2HCl \to CaCl_2 + H_2O$
Theo PTHH :
$n_{HCl} = 2n_{CaO} = 0,2(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$
a, \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Fe}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)=3360\left(ml\right)\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,3\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{FeCl_2}}=\dfrac{0,15}{0,2}=0,75\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\end{matrix}\right.\)
c, Ta có: \(m_{ddHCl}=1,25.200=250\left(g\right)\)
⇒ m dd sau pư = 8,4 + 250 - 0,15.2 = 258,1 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,15.127}{258,1}.100\%\approx7,38\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{258.1}.100\%\approx1,41\%\end{matrix}\right.\)
\(n_{Al}=\dfrac{16,2}{27}=0,6\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 0,6........0,9...........0,3........0,9\left(mol\right)\\ a.V_{H_2\left(đktc\right)}=0,9.22,4=20,16\left(l\right)\\ b.C_{MddH_2SO_4}=\dfrac{0,9}{0,5}=1,8\left(M\right)\\ c.C_{MddX}=C_{MddAl_2\left(SO_4\right)_3}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)
nBa(OH)2 = 0.6 mol
nHCl = 2V (mol)
TH1 : Ba(OH)2 dư
Ba(OH)2 + Al + H2O --> Ba(AlO2)2 + 3/2H2
0.12______0.12
=> nBa(OH)2 phản ứng = 0.6 - 0.12 = 0.48 mol
Ba(OH)2 + 2HCl --> BaCl2 + 2H2O
0.48_______0.96
<=> 2V = 0.96
<=> V = 0.48 (l)
TH2: HCl dư
2Al + 6HCl --> 2AlCl3 + 3H2
0.12___0.36
nHCl phản ứng = 2V - 0.36 (mol)
Ba(OH)2 + 2HCl --> BaCl2 + 2H2O
0.6________1.2
<=> 2V - 0.36 = 1.2
<=> V = 0.78 (l)
TH1 làm sai r bạn ơi! Cân bằng sai pt r!
Th1 : Ba(OH)2 dư
Ta có pt: Ba(OH)2+2Al+2H2O\(\rightarrow Ba\left(AlO_2\right)_2+3H_2\)
0,06 0,12
x+0,06=0,6
x=0,54
nHcl =2.0.54=1.08 mol
VHcl=\(\frac{1.08}{2}=0.54l\)