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![](https://rs.olm.vn/images/avt/0.png?1311)
nFe2O3=24/160=0,15(mol)
Fe2O3+3H2--t*->2Fe+3H2O(1)
0,15____0,45____0,3
Fe2O3+2Al--->Al2O3+Fe(2)
0,15___0,3___________0,15
Fe2O3+3CO--->2Fe+3CO2(3)
0,15____0,45___0,3
(1)VH2=0,45.22,4=10,08(l)
mFe=0,3.56=16,8(g)
(2)mAl=0,3.27=8,1(g)
mFe=0,15.56=8,4(g)
(3)VCO=0,45.22,4=10,08(l)
mFe=0,3.56=16,8(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b+c) Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,15\left(mol\right)\\n_{HCl}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{ddHCl}=\dfrac{0,3\cdot36,5}{10,95\%}=100\left(g\right)\end{matrix}\right.\)
d) PTHH: \(H_2+CuO\xrightarrow[]{t^o}Cu+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=0,15\left(mol\right)\\n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) CuO còn dư
\(\Rightarrow n_{CuO\left(dư\right)}=0,15\left(mol\right)\) \(\Rightarrow m_{CuO\left(dư\right)}=0,15\cdot80=12\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(2Al + 6HCl \to 2AlCl_3 + 3H_2 \)
b)
Tỉ lệ số nguyên tử Al : số phân tử HCl : số phân tử AlCl3 : số phân tử H2 là 2 : 6 : 2 : 3
c)
Ta có : \(n_{Al} = \dfrac{13,5}{27} = 0,5(mol)\)
Theo PTHH : \(n_{H_2} = \dfrac{3}{2}n_{Al} = 0,75(mol)\\ \Rightarrow V_{H_2} = 0,75.22,4 = 16,8(lít)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,4------------>0,4---->0,6
=> \(V_{H_2}=0,6.22,4=13,44\left(l\right)\)
b)
\(m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)
c)
PTHH: CuO + H2 --to--> Cu + H2O
0,6------>0,6
=> mCu = 0,6.64 = 38,4 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.2.......0.4...................0.2\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(C_{M_{HCl}}=\dfrac{0.4}{0.4}=1\left(M\right)\)
\(n_{CuO}=\dfrac{8}{80}=0.1\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(1............1\)
\(0.1.........0.2\)
\(LTL:\dfrac{0.1}{1}< \dfrac{0.2}{1}\Rightarrow H_2dư\)
\(n_{Cu}=n_{CuO}=0.1\left(mol\right)\)
\(m_{Cu}=0.1\cdot64=6.4\left(g\right)\)
Chúc em học tốt và có những trải nghiệm tuyệt vời tại hoc24.vn nhé !
a) nFe=0,2(mol)
PTHH: Fe + 2HCl -> FeCl2 + H2
0,2_________0,4____0,2___0,2(mol)
V(H2,dktc)=0,2.22,4=4,48(l)
b) VddHCl=0,4/0,4=1(l)
c) nCuO=0,1(mol)
PTHH: CuO + H2 -to-> Cu + H2O
Ta có: 0,2/1 > 0,1/1
=> CuO hết, H2 dư, tính theo nCuO
-> nCu=nCuO=0,1(mol)
=>mCu=0,1.64=6,4(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)
\(PTHH:2Al+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,4 <--- 0,6 -----------> 0,2 --> 0,6
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4.27=10,8\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\\V_{H_2}=0,6.22,4=13,44\left(lít\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) nFe= 0,25(mol)
PTHH: Fe + H2SO4 -> FeSO4 + H2
0,25______0,25______0,25__0,25(mol)
b) V(H2,đktc)=0,25.22,4=5,6(l)
c) mH2SO4= 0,25.98= 24,5(g)
Giup mình với, tks u