3.
a. \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,2 0,2 0,2 0,2
Cu không td với \(H_2SO_4\) loãng.
b. \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
=> \(\%_{m_{Fe}}=\dfrac{0,2.56.100}{22,4}=50\%\)
=> \(\%_{m_{Cu}}=100-50=50\%\)
c. \(m_{dd.H_2SO_4.pứ}=\dfrac{0,2.98.100}{9,8}=200\left(g\right)\)
4.
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,3 0,6 0,3 0,3
rắn A: Cu (không tan)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
a. \(m_{Mg}=0,3.24=7,2\left(g\right)\)
=> \(m_{Cu}=20-7,2=12,8\left(g\right)\)
b. \(m_{dd.HCl.tham.gia}=\dfrac{0,6.36,5.100}{10}=219\left(g\right)\)
=> \(V_{dd.HCl.tham.gia}=\dfrac{219}{1,2}=182,5\left(ml\right)\)
c. \(Cu+2H_2SO_{4.đn}\rightarrow CuSO_4+SO_2+2H_2O\)
