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a) $Fe + 2HCl \to FeCl_2 + H_2$
b) Theo PTHH : $n_{FeCl_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$m_{FeCl_2} = 0,2.127 = 25,4(gam)$
c) $n_{H_2} = n_{Fe} = 0,2(mol)$
$V_{H_2} = 0,2.24,79 = 4,958(lít)$
d) $RO + H_2 \xrightarrow{t^o} R + H_2O$
Theo PTHH : $n_{RO} = n_{H_2} = 0,2(mol)$
$\Rightarrow M_{RO} = R + 16 = \dfrac{16}{0,2} = 80$
$\Rightarrow R = 64(Cu)$
CTHH oxit : $CuO$
$n_{Cu} = n_{H_2} = 0,2(mol) \Rightarrow m_{Cu} = 0,2.64 = 12,8(gam)$
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\), ta được CuO dư.
Theo PT: \(n_{Cu}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)
a, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
b, Ta có: \(n_{KMnO_4}=\dfrac{3,16}{158}=0,02\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,01\left(mol\right)\)
\(\Rightarrow m_{O_2}=0,01.32=0,32\left(g\right)\)
c, \(V_{O_2}=0,01.24,79=0,2479\left(l\right)\)
a, \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
b, Theo PT: \(n_{FeSO_4}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{FeSO_4}=0,15.152=22,8\left(g\right)\)
c, \(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,15}{1}\), ta được CuO dư.
Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow n_{CuO\left(dư\right)}=0,3-0,15=0,15\left(mol\right)\)
Chất rắn thu được sau pư gồm Cu và CuO dư.
⇒ m chất rắn = mCu + mCuO (dư) = 0,15.64 + 0,15.80 = 21,6 (g)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
______0,2_________________0,2 (mol)
b, VH2 = 0,2.22,4 = 4,48 (l)
c, Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)
PT: \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\), ta được H2 dư.
Theo PT: \(n_{Fe}=n_{FeO}=0,1\left(mol\right)\)
⇒ mFe = 0,1.56 = 5,6 (g)
Bạn tham khảo nhé!
a) Zn + 2HCl \(\rightarrow\)ZnCl2 + H2
b) mZn = \(\dfrac{13}{65}\)=0,2 (mol)
Zn + 2HCl \(\rightarrow\)ZnCl2 + H2
(mol) 0,2 ----------------------> 0,2
\(V_{H_2}\)= 0,2 . 22,4 = 4,48(lít)
c)\(n_{FeO}\)=\(\dfrac{7,2}{72}\)=0,1 (mol)
H2 + FeO \(\underrightarrow{t^o}\)Fe + H2O
(mol) 0,1----->0,1
mFe = 0,1 . 56 = 5,6(g)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 6HCl ---> 2AlCl3 + 3H2
0,1 0,15
=> VH2 = 0,15.22,4 = 3,36 (l)
\(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
LTL: \(0,4>0,15\rightarrow\) CuO dư
Theo pthh: \(n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,15\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,15.64}{0,15.64+\left(0,4-0,15\right).80}=32,43\%\\\%m_{CuO}=100\%-32,43\%=67,57\%\end{matrix}\right.\)
a. \(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\)
b. \(n_{Al}=\frac{m}{M}=\frac{2,7}{27}=0,1mol\)
Theo phương trình `(1)` \(n_{H_2}=\frac{3}{2}.n_{Al}=\frac{3}{2}.0,1=0,15mol\)
\(\rightarrow V_{H_2\left(ĐKTC\right)}=n.22,4=0,15.22,4=3,36l\)
c. \(CuO+H_2\rightarrow^{t^o}Cu+H_2O\left(2\right)\)
\(n_{CuO}=\frac{m}{M}=\frac{32}{80}=0,4mol\)
Tỷ lệ \(\frac{0,4}{1}>\frac{0,15}{1}\)
`->CuO` dư
Theo phương trình `(2)` \(n_{Cu}=n_{H_2}=0,15mol\)
\(n_{CuO\left(pứ\right)}=n_{H_2}=0,15mol\)
\(\rightarrow n_{CuO\left(dư\right)}=0,4-0,15=0,25mol\)
\(m\left(g\right)\text{ chất rắn }\hept{\begin{cases}CuO_{dư}=0,25mol\\Cu=0,15mol\end{cases}}\)
\(\rightarrow m=0,15.64+0,25.80=29,6g\)
\(\%m_{CuO\left(dư\right)}=\frac{0,25.80.100}{29,6}\approx67,6\%\)
\(\%m_{Cu}=100\%-67,6\%=32,4\%\)
\(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.05................................0.05\)
\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(n_{CuO}=\dfrac{6}{80}=0.075\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(1............1\)
\(0.075......0.05\)
Chất khử : H2 . Chất OXH : CuO
\(LTL:\dfrac{0.075}{1}>\dfrac{0.05}{1}\Rightarrow CuOdư\)
\(m_{CuO\left(dư\right)}=\left(0.075-0.05\right)\cdot64=1.6\left(g\right)\)
a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
____0,1_____0,2______0,1_____0,1 (mol)
\(V_{H_2}=0,1.24,79=2,479\left(l\right)\)
\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
b, \(C_{M_{HCl}}=\dfrac{0,2}{0,1}=1\left(M\right)\)
c, \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\), ta được CuO dư.
Theo PT: \(n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,1\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,05\left(mol\right)\)
⇒ m chất rắn = mCuO (dư) + mCu = 0,05.80 + 0,1.64 = 10,4 (g)
