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Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a. PTHH:
Mg + 2HCl ---> MgCl2 + H2 (1)
MgO + 2HCl ---> MgCl2 + H2O (2)
Theo PT(1): \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
=> \(m_{Mg}=0,1.24=2,4\left(g\right)\)
=> \(m_{MgO}=4,4-2,4=2\left(g\right)\)
b. Ta có: \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
=> \(n_{hh}=0,05+0,1=0,15\left(mol\right)\)
Theo PT(1,2): \(n_{HCl}=2.n_{hh}=2.0,15=0,3\left(mol\right)\)
=> \(V_{dd_{HCl}}=\dfrac{0,3}{0,4}=0,75\left(lít\right)\)
a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\) (*)
Phương trình hóa học
Mg + 2HCl ---> MgCl2 + H2 (**)
MgO + 2HCl ---> MgCl2 + H2O (***)
b) Từ (*) và (**) ta có \(n_{Mg}=0,15\Leftrightarrow m_{Mg}=0,15.24=3,6\left(g\right)\)
\(\Rightarrow m_{MgO}=10-3,6=6,4\left(g\right)\)
\(\%Mg=\dfrac{3,6}{10}.100\%=36\%\)
\(\%MgO=\dfrac{6,4}{10}.100\%=64\%\)
c) Xét phản ứng (**) ta có \(m_{MgO}=6,4\left(g\right)\Leftrightarrow n_{MgO}=n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,16\left(mol\right)\) (1)
\(\Leftrightarrow n_{HCl}=0,32\left(mol\right)\)
Tương tự có số mol HCl trong phản ứng (*) là 0,3 mol
\(C_M=\dfrac{0,32+0,3}{0,2}=3,1\left(M\right)\)
d) Từ (1) ; (*) ; (**) ta có : \(n_{MgCl_2}=0,15+0,16=0,31\left(mol\right)\)
\(m_{MgCl_2}=0,31.95=29,45\left(g\right)\)
e) \(C_M=\dfrac{0,31}{0,2}=1,55\left(M\right)\)
a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
_____0,2<---0,6<--------------0,3
=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,2.27}{15,6}.100\%=34,615\%\\\%Al_2O_3=\dfrac{15,6-0,2.27}{15,6}.100\%=65,385\%\end{matrix}\right.\)
b) \(n_{Al_2O_3}=\dfrac{15,6-0,2.27}{102}=0,1\left(mol\right)\)
PTHH: Al2O3 + 6HCl --> 2AlCl3 + 3H2O
______0,1--->0,6
=> nHCl = 0,6+0,6 = 1,2(mol)
=> \(V_{dd}=\dfrac{1,2}{2}=0,6\left(l\right)\)
2CH3COOH+CaCO3-to>(CH3COO)2Ca+H2O+CO2
0,4-----------------0,2----------------------------------------0,2
2CH3COOH+CaO->(CH3COO)2Ca+H2O
0,1----------------0,05
n CO2=0,2 mol
=>%m CaCO3=\(\dfrac{0,2.100}{22,8}100=87,72\%\)
=>%m CaO=12,28%
=>n CaO=0,05 mol
=>VCH3COOH=\(\dfrac{0,5}{2}=0,25l\)
a)
\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: CaCO3 + 2CH3COOH --> (CH3COO)2Ca + CO2 + H2O
0,2<---------0,4<------------------------------0,2
=> \(m_{CaCO_3}=0,2.100=20\left(g\right)\)
=> \(\left\{{}\begin{matrix}\%m_{CaCO_3}=\dfrac{20}{22,8}=87,72\%\\\%m_{CaO}=100\%-87,72\%=12,28\%\end{matrix}\right.\)
b)
\(n_{CaO}=\dfrac{22,8-20}{56}=0,05\left(mol\right)\)
PTHH: CaO + 2CH3COOH --> (CH3COO)2Ca + H2O
0,05---->0,1
=> \(V_{dd.CH_3COOH}=\dfrac{0,1+0,4}{2}=0,25\left(l\right)\)
c) \(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{a}{60}\left(mol\right)\\n_{C_2H_5OH}=\dfrac{1,5a}{46}\left(mol\right)\\n_{CH_3COOC_2H_5}=\dfrac{1,2a}{88}\left(mol\right)\end{matrix}\right.\)
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
Xét tỉ lệ: \(\dfrac{\dfrac{a}{60}}{1}< \dfrac{\dfrac{1,5a}{46}}{1}\) => HIệu suất tính theo CH3COOH
\(n_{CH_3COOH\left(pư\right)}=\dfrac{1,2a}{88}\left(mol\right)\)
=> \(H=\dfrac{\dfrac{1,2a}{88}}{\dfrac{a}{60}}.100\%=81,82\%\)
a)\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,15 0,3 0,15
\(m_{Zn}=0,15\cdot65=9,75\left(g\right)\)
\(\%m_{Zn}=\dfrac{9,75}{17,85}\cdot100\%=54,62\%\)
\(\%m_{ZnO}=100\%-54,62\%=45,38\%\)
b)\(m_{ZnO}=17,85-9,75=8,1\left(g\right)\Rightarrow n_{ZnO}=\dfrac{8,1}{81}=0,1mol\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
0,1 0,2
\(\Rightarrow\Sigma n_{HCl}=0,3+0,2=0,5mol\)
\(\Rightarrow V=\dfrac{0,5}{1}=0,5l=500ml\)
a
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,15<-0,15<--0,15<----0,15
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,16-->0,32---->0,16
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(m_{Mg}=0,15.24=3,6\left(g\right)\\ m_{MgO}=10-3,6=6,4\left(g\right)\)
b
\(\%m_{Mg}=\dfrac{3,6.100\%}{10}=36\%\\ \%m_{MgO}=\dfrac{6,4.100\%}{10}=64\%\)
c
\(n_{MgO}=\dfrac{6,4}{40}=0,16\left(mol\right)\)
\(CM_{HCl}=\dfrac{0,15+0,32}{0,2}=2,35M\)
d
\(m_{MgCl_2}=\left(0,15+0,16\right).95=29,45\left(g\right)\)
e
\(CM_{MgCl_2}=\dfrac{0,15+0,16}{0,2}=1,55M\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Mg}\)
\(\Rightarrow m_{Mg}=0,1\cdot24=2,4\left(g\right)\) \(\Rightarrow m_{MgO}=2\left(g\right)\)
a, Ta có nhỗn hợp khí A = \(\dfrac{3,36}{22,4}\) = 0,15 ( mol )
Mg + 2HCl \(\rightarrow\) MgCl2 + H2
x \(\rightarrow\) 2x \(\rightarrow\) x \(\rightarrow\) x
CaCO3 + 2HCl \(\rightarrow\) CaCl2 + H2O + CO2
y \(\rightarrow\) 2y \(\rightarrow\) y \(\rightarrow\) y \(\rightarrow\) y
=> \(\left\{{}\begin{matrix}24x+100y=7,4\\x+2y=0,15\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\dfrac{1}{260}\\y=\dfrac{19}{260}\end{matrix}\right.\)
=> mMg = 24 . \(\dfrac{1}{260}\) = \(\dfrac{6}{65}\) ( gam )
=> mCaCO3 = 100 . \(\dfrac{19}{260}\) = \(\dfrac{95}{13}\) ( gam )
Câu a mình giải sai chỗ lập hệ phương trình
5 dòng đầu đúng rồi mình xin làm lại các dòng sau
=> \(\left\{{}\begin{matrix}24x+100y=7,4\\x+y=0,15\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
=> mMg = 24 . 0,1 = 2,4 ( gam )
=> mCaCO3 = 7,4 - 2,4 = 5 ( gam )