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\(n_{CuO}=\dfrac{16}{80}=0,2mol\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,2 0,4 0,2 0,2
\(C_{M_{HCl}}=\dfrac{n_{HCl}}{V_{HCl}}=\dfrac{0,4}{0,2}=2M\)
\(m_{HCl}=0,4\cdot36,5=14,6g\)
nCuO = 16/80 = 0,2 (mol)
PTHH: CuO + 2HCl -> CuCl2 + H2
Mol: 0,2 ---> 0,4 ---> 0,2 ---> 0,2
CMCuCl2 = 0,2/0,2 = 1M
mHCl = 0,4 . 36,5 = 14,6 (g)
\(a.300ml=0,3l\\ n_{Ba\left(OH\right)_2}=0,3.0,5=0,15mol\\ Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
0,15 0,3 0,15
\(200ml=0,2l\\ C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5M\\ b.C_{M_{BaCl_2}}=\dfrac{0,15}{0,3+0,2}=0,3M\)
\(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,05
b) \(n_{H2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
c) \(n_{HCl}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)
200ml = 0,2l
\(C_{M_{ddHCl}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
Chúc bạn học tốt
nMg = 6,72 : 22,4 = 0,3 mol
Mg + 2HCl -> MgCl2 + H2
0,3 0,6 0,3
=> mMg = 0,3 . 24 = 7,2 g
CM HCl = 0,6 : 0,5 = 4M
\(a,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2-->0,4----------------->0,2
\(maxV_{H_2}=0,2.22,4=4,48\left(l\right)\\ b,C_{M\left(HCl\right)}=\dfrac{0,4}{0,2}=2M\\ c,n_{Cl}=m_{HCl}=0,4\left(mol\right)\\ BTKL:m_{muối}=m_{KL}+m_{Cl}=5,5+0,4.35,5=19,7\left(g\right)\)
\(V_{\text{dd}}=0,2+0,3=0,5\left(l\right)\\ n_{HCl}=0,2.1+0,3.0,5=0,35\\ C_M=\dfrac{0,35}{0,5}=0,7M\)
200ml = 0,2(l)
=> nHCl (1) = 0,2 .1 = 0,2 (mol)
300ml = 0,3 (l)
=> nHCl(2) = 0,3 . 0,5 = 0,15 (mol)
=> CM (sau khi trộn) = n/V = (0,15+0,2) / (0,2+0,3 ) = 0,35 / 0,5 = 0,7 M
mol MG=24\24=1mol
MG +2HCL _____ MGCL2 +H2
pt:1mol 2mol
đb:1mol ?
theo pt : molHCL = 2mol
300mi=0.3l
Cm= n\v= 2/0.3=6.7mol/l
Số mol Mg \(n_{Mg}=\frac{24}{24}=1\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
1 2 (mol)
Nồng độ mol của dd HCl: \(C_{M\left(ddHCl\right)}=\frac{2}{0,3}=6,67M\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,1---->0,1----------------->0,1
=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\end{matrix}\right.\)
PTHH: CuO + H2 --to--> Cu + H2O
LTL: 0,25 > 0,1 => CuO dư
Theo pthh: nCu = nH2 = 0,1 (mol)
=> mCu = 0,1.64 = 6,4 (g)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,2
\(V_{H_2}=0,2.22,4=4,48l\\
C_M=\dfrac{0,2}{0,2}=1M\\
n_{CuO}=\dfrac{20}{80}=0,25\left(G\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\
LTL:0,25>0,1\)
=>CuO dư
\(n_{Cu}=n_{H_2}=0,1\left(mol\right)\\
m_{Cu}=0,1.64=6,4g\)
\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(0.2.......0.4........................0.2\)
\(C_{M_{HCl}}=\dfrac{0.4}{0.2}=2\left(M\right)\)
\(n_{CuO}=\dfrac{32}{80}=0.4\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
Lập tỉ lệ : \(\dfrac{0.4}{1}>\dfrac{0.2}{1}\)
=> CuO dư
\(m_{cr}=m_{CuO\left(dư\right)}+m_{Cu}=32-0.2\cdot80+0.2\cdot64=28.8\left(g\right)\)
\(\%Cu=\dfrac{0.2\cdot64}{28.8}\cdot100\%=44.44\%\)
\(\%CuO\left(dư\right)=55.56\%\)
\(V_{HCl}=200ml=0,2l\\ n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\\ PTHH:Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\\ n_{HCl}=0,25.2=0,5\left(mol\right)\\ C_{MHCl}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)
\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PT: \(n_{MgCl_2}=n_{Mg}=0,25\left(mol\right)\)
\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)