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CuO + 2HCl ------> CuCl2 + H2O
nHCl bđ = 0.5.1.4 = 0.7
nCuO = \(\dfrac{16}{80}\)= 0.2 mol
=> nHCl pư = 0.4 mol
=> nHCl dư = 0.3 mol
Dung dịch A gồm CuCl2 và HCl dư
mCuCl2 = 0.2.135 = 27g
mHCl dư = 0.3.36.5 = 10.95g
CM CuCl2 = \(\dfrac{0,2}{0,5}\)= 0.4M
CM HCl dư =\(\dfrac{0,3}{0,5}\) = 0.6M
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
a, Ta có:
\(n_{H2}=\frac{0,224}{22,4}=0,01\left(mol\right)\)
\(\Rightarrow n_{Mg}=0,01\left(mol\right)\)
\(\%m_{Mg}=\frac{0,01.24}{1,04}.100\%=23,08\%\)
\(\%m_{CuO}=100\%-23,08\%=76,92\%\)
b,\(n_{CuO}=\frac{0,8}{80}=0,01\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,01.2+0,01.2=0,04\)
\(CM_{HCl}=\frac{0,04}{0,25}=0,16M\)
\(\Rightarrow m=m_{MgCl2}+m_{CuCl2}=0,01.95+0,01.135=2,3\left(g\right)\)
\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)
\(n_{HCl}=0,1.0,3=0,03\left(mol\right)\)
=> \(n_{H_2O}=\dfrac{0,03}{2}=0,015\left(mol\right)\)
Theo ĐLBTKL: \(m_{oxit}+m_{HCl}=m_{muối}+m_{H_2O}\)
=> mmuối = 3,425 + 0,03.36,5 - 0,015.18 = 4,25(g)
Bài 1:
\(n_{HCl}=2.0,16=0,32\left(mol\right);n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
PTHH: Fe + 2HCl → FeCl2 + H2
\(m_{H_2}=0,16.2=0,32\left(g\right)\)
\(m_{HCl}=0,32.36,5=11,68\left(g\right)\)
Theo ĐLBTKL ta có: \(m_{MgCl_2+FeCl_2}=1,4+11,68-0,32=12,76\left(g\right)\)
Bài 12:
Theo ĐLBTKL, ta có:
\(m_{hhkl}+m_{O_2}=m_{hh.oxit}\\ \Leftrightarrow11,9+m_{O_2}=18,3\\ \Leftrightarrow m_{O_2}=18,3-11,9=6,4\left(g\right)\\ n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
a) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_S=\dfrac{1,6}{32}=0,05\left(mol\right)\)
PTHH: Fe + S --to--> FeS
0,05<-0,05-->0,05
Fe + 2HCl --> FeCl2 + H2
0,05->0,1---->0,05-->0,05
FeS + 2HCl --> FeCl2 + H2S
0,05-->0,1----->0,05--->0,05
=> \(\%V_{H_2S}=\%V_{H_2}=\dfrac{0,05}{0,05+0,05}.100\%=50\%\)
b)
nNaOH = 0,3 (mol)
- Gọi số mol HCl trong B là a (mol)
PTHH: NaOH + HCl --> NaCl + H2O
a<-----a
FeCl2 + 2NaOH --> Fe(OH)2 + 2NaCl
0,1---->0,2
=> a + 0,2 = 0,3
=> a = 0,1 (mol)
\(C_{M\left(FeCl_2\left(B\right)\right)}=\dfrac{0,1}{0,5}=0,2M\)
\(C_{M\left(HCl\left(B\right)\right)}=\dfrac{0,1}{0,5}=0,2M\)
nHCl(bđ) = 0,3 (mol)
=> \(C_{M\left(dd.HCl\left(bđ\right)\right)}=\dfrac{0,3}{0,5}=0,6M\)
\(a,PTHH:X+2HCl\to XCl_2+H_2\\ \Rightarrow n_{X}=n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \Rightarrow M_X=\dfrac{9,75}{0,15}=65(g/mol)(Zn)\\ b,n_{HCl}=2.0,2=0,4(mol)\)
Vì \(\dfrac{n_{H_2}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư
\(\Rightarrow n_{ZnCl_2}=n_{H_2}=0,15(mol)\\ \Rightarrow m_{ZnCl_2}=136.0,15=20,4(g)\\ C_{M_{ZnCl_2}}=\dfrac{0,15}{0,2}=0,75M\)
\(n_{HCl}=1,4.0,5=0,7\left(mol\right)\)
\(n_{CuO}=\frac{16}{80}=0,2\left(mol\right)\)
\(PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\)
(mol)_____0,2_____0,4_____0,2___________
Tỉ lệ: \(\frac{0,7}{2}>\frac{0,2}{1}\rightarrow HCl\) dư
\(m_{CuCl_2}=0,2.135=27\left(g\right)\)