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Câu 1 :\(n_{CO_2} = \dfrac{2,688}{22,4} = 0,12(mol)\)
MgCO3 + 2HCl \(\to\) MgCl2 + CO2 + H2O
..................................0,12........0,12..................(mol)
Suy ra: a = 0,12.95 = 11,4(gam)
Câu 2 :
\(Fe + 2HCl \to FeCl_2 + H_2\\ n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ \Rightarrow n_{Cu} = 2n_{Fe} = 0,15.2 = 0,3(mol)\\ 2Fe+3Cl_2\xrightarrow{t^o} 2FeCl_3\\ Cu+Cl_2 \xrightarrow{t^o} CuCl_2\\ n_{Cl_2} = \dfrac{3}{2}n_{Fe} + n_{Cu} = 0,525\\ \Rightarrow V = 0,525.22,4 =11,76(lít)\)
n H2=\(\dfrac{1,12}{22,4}\)=0,05 mol
Zn+2HCl->ZnCl2+H2
0,05---0,1-----0,05---------0,05 mol
ZnO+2HCl->ZnCl2+H2
0,07----0,14---0,07
=m Zn=0,05.65=3,25g
m ZnCl2=0,05.136=6,8g
=>m ZnCl2 pt2 =16,32-6,8=9,52g
=>n ZnCl2=\(\dfrac{9,52}{136}\)=0,07 mol
=>m =3,25+0,07.81=8,92g
=>VHCl=\(\dfrac{0,24}{0,5}\)=0,48l=480ml
cÂU 2.
\(n_Z=\dfrac{6,72}{22,4}=0,3mol\)
\(\left\{{}\begin{matrix}n_{CaCO_3}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\Rightarrow100x+56y=25,6\left(1\right)\)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\Rightarrow x+y=n_Z=0,3\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\%m_{CaCO_3}=\dfrac{0,2\cdot100}{25,6}\cdot100\%=78,125\%\)
\(\%m_{Fe}=100\%-78,125\%=21,875\%\)
\(m_{muối}=m_{CaCl_2}+m_{FeCl_2}=0,2\cdot111+0,1\cdot127=34,9g\)
-
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
Mg + 2HCl --> MgCl2 + H2
-
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{200.18,25}{100.36,5}=1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,2<----0,4<---------------0,2
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
0,1<-----0,6
=> \(\left\{{}\begin{matrix}\%Mg=\dfrac{0,2.24}{0,2.24+0,1.160}.100\%=23,077\%\\\%Fe_2O_3=\dfrac{0,1.160}{0,2.24+0,1.160}.100\%=76,923\%\end{matrix}\right.\)
Đáp án B
Khí thi được là
Cu là kim loại đứng sau hiđro trong dãy hoạt động hóa học, do đó Cu không tác dụng với dung dịch H 2 SO 4 loãng => Chất rắn không tan là Cu
Sơ đồ phản ứng:
a.\(Fe+S\rightarrow\left(t^o\right)FeS\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
b.\(n_{hhk}=\dfrac{4,48}{22,4}=0,2mol\)
\(Fe+S\rightarrow\left(t^o\right)FeS\)
Ta thu được hh khí --> S hết, Fe dư
Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_S=y\end{matrix}\right.\)
\(\rightarrow n_{FeS}=n_{Fe}=n_S\rightarrow n_{Fe\left(dư\right)}=x-y\) ( mol )
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(x-y\) \(x-y\) ( mol )
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
y y ( mol )
Ta có: \(\left(x-y\right)+y=0,2\)
\(\Leftrightarrow x=0,2\)
Ta có:\(56x+32y=14,4\)
\(\Leftrightarrow56.0,2+32y=14,4\)
\(\Leftrightarrow y=0,1\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2.56}{14,4}.100=77,77\%\\\%m_S=100\%-77,77\%=22,23\%\end{matrix}\right.\)
\(Đặt:\)
\(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
\(m_{hh}=24x+56y=13.6\left(g\right)\\ n_{H_2}=x+y=0.3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x=0.1\\y=0.2\end{matrix}\right.\)
\(\%Mg=\dfrac{0.1\cdot24}{13.6}\cdot100\%=17.64\%\\ \%Fe=100-17.64=82.36\%\)
\(n_{HCl}=2n_{H_2}=2\cdot0.3=0.6\left(mol\right)\)
\(V_{HCl}=\dfrac{0.6}{2}=0.3\left(l\right)\)
\(m_Y=m_{MgCl_2}+m_{FeCl_2}=0.1\cdot95+0.2\cdot127=34.9\left(g\right)\)
\(n_{HCl}=\dfrac{100.18,25}{36,5.100}=0,5mol\)
\(n_{H_2}=0,15mol\)
Zn+2HCl\(\rightarrow\)ZnCl2+H2(1)
ZnO+2HCl\(\rightarrow\)ZnCl2+H2O(2)
\(n_{ZnCl_2}=\dfrac{27,2}{136}=0,2mol\)
- Theo PTHH (1): \(n_{Zn}=n_{ZnCl_2\left(1\right)}=n_{H_2}=0,15mol\)
- Theo PTHH (2):\(n_{ZnO}=n_{ZnCl_2\left(2\right)}=0,2-0,15=0,05mol\)
- Theo PTHH (1,2): \(n_{HCl\left(pu\right)}=2n_{ZnCl_2}=2.0,2=0,4mol\)
\(n_{HCl\left(dư\right)}=0,5-0,4=0,1mol\)
m=\(m_{Zn}+m_{ZnO}=0,15.65+0,05.81=13,8g\)
\(m_{dd}=13,8+100-0,15.2=113,5g\)
\(C\%_{HCl}=\dfrac{0,1.36,5.100}{113,5}\approx3,22\%\)
\(C\%_{ZnCl_2}=\dfrac{27,2.100}{113,5}\approx24\%\)