a) \(Mg+FeCl_2\rightarrow MgCl_2+Fe\)
b) Gọi x là số mol Mg phản ứng
Ta có : \(m_{tăng}=m_{Fe}-m_{Mg}=56x-24x=36,8-24=12,8\)
=> x=0,4(mol)
\(\Rightarrow m_{Mg\left(pứ\right)}=0,4.24=9,6\left(g\right);m_{Fe}=0,4.56=22,4\left(g\right)\)
c) \(\%m_{Fe}=\dfrac{22,4}{36,8}.100=60,87\%\); \(\%m_{Mg}=100-60,87=39,13\%\)
d) \(m_{Mg\left(saupu\right)}=36,8-22,4=14,4\left(g\right)\Rightarrow n_{Mg\left(saupu\right)}=0,6\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,6------------------------------------->0,6
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,4------------------------------------->0,4
\(n_{HCl}=0,6+0,4=1\left(mol\:\right)\Rightarrow V_{HCl}=1.22,4=22,4\left(l\right)\)
e) Do tác dụng với HNO3 đặc nên khí thoát ra là NO2
\(Mg\rightarrow Mg^{2+}+2e\) \(N^{+5}+2e\rightarrow N^{+4}\)
\(Fe\rightarrow Fe^{3+}+3e\)
\(BTe:n_{Mg}.2+n_{Fe}.3=n_{NO_2}.1\)
=> \(n_{NO_2}=2,4\left(mol\right)\Rightarrow V_{NO_2}=2,4.22,4=53,76\left(l\right)\)
