Question

bỏ Mg vào 1,2l dd  Sắt(III) chloride 0,1M. sau pư ht thu đc 0.00336 kg kết tủa. kl Mg là

Answer 1

\(n_{FeCl_3}=0,1.1,2=0,12\left(mol\right)\\ Mg+2FeCl_3\rightarrow2MgCl_2+FeCl_2\\ Mg_{dư}+FeCl_2\rightarrow MgCl_2+Fe\\ m_{kết.tủa}=0,00336\left(kg\right)=3,36\left(g\right)=m_{Fe}\\ \Rightarrow n_{Mg\left(dư\right)}=n_{Fe}=\dfrac{3,36}{56}=0,06\left(mol\right)\\ n_{Mg\left(dùng\right)}=\dfrac{n_{FeCl_3}}{2}+n_{Mg\left(dư\right)}=\dfrac{0,12}{2}+0,06=0,12\left(mol\right)\\ \Rightarrow m_{Mg\left(dùng\right)}=0,12.24=2,88\left(g\right)\)