4.\(n_{H_2}=\dfrac{1,008}{22,4}=0,045\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có \(n_{Fe}=n_{H_2}=0,045\left(mol\right)\)
=> \(m_{Fe}=0,045.56=2,52\left(g\right)\)
=> \(m_{Ag}=4,68-2,52=2,16\left(g\right)\)
b) \(\%m_{Al}=\dfrac{2,52}{4,68}.100=53,85\%\)
=> %mCu = 100- 53,85 = 46,15%
c) \(n_{HCl}=2n_{H_2}=0,09\left(mol\right)\)
\(V_{HCl}=\dfrac{0,09}{2}=0,045\left(l\right)\)
1.a) \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Ta có : \(n_{Mg}=n_{H2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\)
=> \(m_{Mg}=0,07.24=1,68\left(g\right)\)
=> \(m_{Cu}=3-1,68=1,32\left(g\right)\)
b) \(\%m_{Mg}=\dfrac{1,68}{3}.100=56\%\)
=> \(\%m_{Cu}=100-56=44\%\)
c)\(n_{H_2SO_4}=n_{H_2}=0,07\left(mol\right)\)
=>\(CM_{H_2SO_4}=\dfrac{0,07}{0,1}=0,7M\)
2.\(n_{H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Ta có \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,01\left(mol\right)\)
=> \(m_{Al}=0,01.27=0,27\left(g\right)\)
=> \(m_{Cu}=0,6-0,27=0,33\left(g\right)\)
b) \(\%m_{Al}=\dfrac{0,27}{0,6}.100=45\%\)
=> %mCu = 100-45 = 55%
c) \(n_{HCl}=2n_{H_2}=0,03\left(mol\right)\)
\(V_{HCl}=\dfrac{0,03}{1}=0,03\left(l\right)\)
3.a)\(n_{H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(n_{Fe}=n_{H_2}=0,015\left(mol\right)\)
=> \(m_{Fe}=0,015.56=0,84\left(mol\right)\)
\(m_{Ag}=1,5-0,84=0,66\left(mol\right)\)
b) \(\%m_{Fe}=\dfrac{0,84}{1,5}.100=56\%\)
\(\%m_{Ag}=\dfrac{0,66}{1,5}.100=44\%\)
c) \(n_{H_2SO_4}=n_{H_2}=0,015\left(mol\right)\)
=> \(CM_{H_2SO_4}=\dfrac{0,015}{0,2}=0,075M\)