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\(n_{Fe}=a\left(mol\right),n_{Mg}=b\left(mol\right)\)
\(m_{hh}=56a+24b=10.16\left(g\right)\)
\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(n_{H_2}=a+b=0.25\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.13,b=0.12\)
\(m_{Fe}=0.13\cdot56=7.28\left(g\right)\)
\(m_{Mg}=0.12\cdot24=2.88\left(g\right)\)
\(n_{HCl}=2\cdot n_{H_2}=2\cdot0.25=0.5\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{0.5}{0.5}=1\left(M\right)\)
nH2 = 2.24/22.4 = 0.1 (mol)
Fe + 2HCl => FeCl2 + H2
0.1___0.2_____0.1___0.1
mFeO = 12.8 - 0.1*56 = 7.2 (g)
nFeO = 7.2/72 = 0.1 (mol)
FeO + 2HCl => FeCl2 + H2O
0.1____0.2______0.1
%Fe = 5.6/12.8 * 100% = 43.75%
%FeO = 56.25%
nHCl = 0.2 + 0.2 = 0.4 (mol)
Vdd HCl = 0.4/0.1 = 4(l)
nFeCl2 = 0.1 + 0.1 = 0.2 (mol)
CM FeCl2 = 0.2/4 = 0.05 (M)
1.
a, \(2Fe+3Cl_2\underrightarrow{^{to}}2FeCl_3\)
\(m_{FeCl_3}=\frac{16,25.100}{100}=16,25\left(g\right)\)
b, \(n_{FeCl_3}=\frac{16,25}{162,5}=0,1\left(mol\right)\)
\(\rightarrow n_{Fe}=n_{FeCl_3}=0,1\left(mol\right)\)
\(\rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(n_{CL2}=\frac{3}{2}n_{FeCl3}=0,15\left(mol\right)\)
\(\rightarrow V_{CL2}=0,15.22,4=3,36\left(l\right)\)
Bài 2 :
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
\(n_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(\rightarrow n_{Zn}=n_{H2}=0,1\left(mol\right)\)
\(m_{Zn}=0,1.65=6,5\left(g\right),m_{ZnO}=10,55-6,5=4,05\left(g\right)\)
b)
\(n_{ZnO}=\frac{4,05}{81}=0,05\left(mol\right)\)
\(n_{HCl}=0,05.2+0,1.2=0,3\left(mol\right)\)
\(\rightarrow m_{dd_{HCl}}=\frac{0,3.36,5}{10\%}=109,5\left(g\right)\)
Bài 3 : Xem lại đề
Bài 4:
a)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{HCl}=0,25.2=0,5\left(mol\right)\)
Gọi a là số mol Fe b là số mol Zn
Giải hệ phương trình :
\(\left\{{}\begin{matrix}56a+65b=14,9\\2a+2b=0,5\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\end{matrix}\right.\)
\(\%m_{Fe}=\frac{0,15.56}{14,9}.100\%=56,38\%,\%m_{Zn}=100\%-56,38\%=43,62\%\)
b)
\(n_{H2}=\frac{n_{HCl}}{2}=\frac{0,5}{2}=0,25\left(mol\right)\)
\(\rightarrow V_{H2}=0,25.22,4=5,6\left(l\right)\)
Bài 5 :
m tăng thêm=mKl-mH2
\(\rightarrow m_{H2}=7,8-7=0,8\left(g\right)\)
\(\rightarrow n_{H2}=\frac{0,8}{2}=0,4\left(mol\right)\)
Gọi a là số mol Al b là số mol Mg
Giải hệ phương trình :
\(\left\{{}\begin{matrix}27a+24b=7,8\\1,5a+b=0,4\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
\(m_{Al}=0,2.27=5,4\left(g\right),m_{Mg}=0,1.24=2,4\left(g\right)\)
