Có [tex]n_{H_{2}}[/tex] = [tex]\frac{2,24}{22,4}[/tex] = 0,1 ( mol )
BTKL : [tex]m_{Cl_{2}}[/tex] = [tex]m_{hh muối}[/tex][tex]-m_{hh KL}[/tex] = 12,526 - 5 = 7,526 (g)
=> [tex]n_{Cl_{2}}[/tex] = 0,106 ( mol )
BT e có: [tex]2n_{Mg}+3n_{Al}+2n_{Zn}+2n_{Fe}=2n_{H_{2}}=0,2[/tex] (mol)
[tex]2n_{Mg}+3n_{Al}+2n_{Zn}+3n_{Fe}=2n_{Cl_{2}}=0,212[/tex] (mol)
=> [tex]n_{Fe}[/tex] = 0,012 (mol)
=> [tex]m_{Fe}[/tex] = 0,012.56 =0,672 (g)