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nAl = 5.4/27 = 0.2 (mol)
2Al + 6HCl => 2AlCl3 + 3H2
0.2.......0.6......................0.3
CM HCl = 0.6 / 0.4 = 1.5 (M)
nCuO = 32/80 = 0.4 (mol)
CuO + H2 -to-> Cu + H2O
0.2.......0.2..........0.2
Chất rắn : 0.2 (mol) CuO dư , 0.2 (mol) Cu
%CuO = 0.2*80 / ( 0.2*80 + 0.2*64) * 100% = 55.56%
%Cu = 44.44%
\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(0.2.......0.4........................0.2\)
\(C_{M_{HCl}}=\dfrac{0.4}{0.2}=2\left(M\right)\)
\(n_{CuO}=\dfrac{32}{80}=0.4\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
Lập tỉ lệ : \(\dfrac{0.4}{1}>\dfrac{0.2}{1}\)
=> CuO dư
\(m_{cr}=m_{CuO\left(dư\right)}+m_{Cu}=32-0.2\cdot80+0.2\cdot64=28.8\left(g\right)\)
\(\%Cu=\dfrac{0.2\cdot64}{28.8}\cdot100\%=44.44\%\)
\(\%CuO\left(dư\right)=55.56\%\)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,2---------------------->0,3
=> VH2 = 0,3.22,4 = 6,72 (l)
b)
\(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,3}{1}\) => CuO dư, H2 hết
PTHH: CuO + H2 --to--> Cu + H2O
0,3<--0,3------->0,3
=> Rắn sau pư gồm \(\left\{{}\begin{matrix}Cu:0,3\left(mol\right)\\CuO\left(dư\right):0,1\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,3.64}{0,3.64+0,1.80}.100\%=70,59\%\\\%m_{CuO}=\dfrac{0,1.80}{0,3.64+0,1.80}.100\%=29,41\%\end{matrix}\right.\)
\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
a)\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,6 0,3
\(C_M=\dfrac{0,6}{0,4}=1,5M\)
b)\(n_{CuO}=\dfrac{32}{80}=0,4mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,4 0,3 0,3
Sau phản ứng CuO dư và dư \(\left(0,4-0,3\right)\cdot80=8g\)
\(m_{rắn}=m_{Cu}=0,3\cdot64=19,2g\)
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.2.........................0.2.......0.3\)
\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)
\(n_{CuO}=\dfrac{32}{160}=0.2\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(1..........1\)
\(0.2........0.3\)
\(LTL:\dfrac{0.2}{1}< \dfrac{0.3}{1}\Rightarrow H_2dư\)
\(n_{Cu}=0.2\left(mol\right)\)
\(m_{Cu}=0.2\cdot64=12.8\left(g\right)\)
Em xem lại đề vì chất rắn chỉ có Cu không có CuO nhé !
a, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Theo PT: \(n_{HCl}=3n_{Al}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,4}=1,5\left(M\right)\)
b, \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PT: \(n_{Cu}=n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Cu}=0,3.64=19,2\left(g\right)\)
a) 2Al+6HCl→2AlCl3+3H22Al+6HCl→2AlCl3+3H2
b) nAl=5,427=0,2(mol)nAl=5,427=0,2(mol)
Theo phương trình : nH2=32nAl=0,3(mol)nH2=32nAl=0,3(mol)
→VH2(đktc)=0,3.22,4=6,72(l)→VH2(đktc)=0,3.22,4=6,72(l)
c) Chất rắn : 0,2(mol)0,2(mol)
CuO dư : 0,2(mol)Cu0,2(mol)Cu
%CuO=0,2.80(0,2.80+0,2.64).100=55,56%%CuO=0,2.80(0,2.80+0,2.64).100=55,56%
%Cu=44,44%%Cu=44,44%
a)\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,1 0,3
b)\(V_{H_2}=0,3\cdot22,4=6,72l\)
c)\(n_{CuO}=\dfrac{32}{80}=0,4mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,4 0,3 0,3
\(m_{Cu}=0,3\cdot64=19,2g\)
nAl = 5.4/27 = 0.2 (mol)
2Al + 6HCl => 2AlCl3 + 3H2
0.2.......0.6......................0.3
CM HCl = 0.6 / 0.4 = 1.5 (M)
nCuO = 32/80 = 0.4 (mol)
CuO + H2 -to-> Cu + H2O
0.2.......0.2..........0.2
Chất rắn : 0.2 (mol) CuO dư , 0.2 (mol) Cu
%CuO =\(\dfrac{0,2.80}{0,2.80+0,2.64}\) 100% = 55.56%
%Cu = 44.44%
a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo phương trình : \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)
\(\rightarrow V_{H_2}\left(đktc\right)=0,3.22,4=6,72\left(l\right)\)
c) Chất rắn : \(0,2\left(mol\right)\)
CuO dư : \(0,2\left(mol\right)Cu\)
\(\%CuO=\dfrac{0,2.80}{\left(0,2.80+0,2.64\right)}.100=55,56\%\)
\(\%Cu=44,44\%\)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
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)` \(n_{Cu}=n_{H_2}=0,15mol\)
\(n_{CuO\left(pứ\right)}=n_{H_2}=0,15mol\)
\(\rightarrow n_{CuO\left(dư\right)}=0,4-0,15=0,25mol\)
\(m\left(g\right)\text{ chất rắn }\hept{\begin{cases}CuO_{dư}=0,25mol\\Cu=0,15mol\end{cases}}\)
\(\rightarrow m=0,15.64+0,25.80=29,6g\)
\(\%m_{CuO\left(dư\right)}=\frac{0,25.80.100}{29,6}\approx67,6\%\)
\(\%m_{Cu}=100\%-67,6\%=32,4\%\)
bn ơi mik có lm r...
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,3 0,45
\(C_{M\left(H_2SO_4\right)}=\dfrac{0,45}{0,3}=1,5M\\ n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\\ pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(LTL:\dfrac{0,5}{1}>\dfrac{0,45}{1}\)
=> CuO dư
\(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,45\left(mol\right)\\ m_{Cr}=\left(0,5-0,45\right).80+0,45.64=32,8g\)