3Mg+2FeCl3->3MgCl2+2Fe
0,375x--0,25x----------------0,25x
=>\(\Delta\)m=0,25x.56-0,375x.24 =1,2
=>x=0,24 M
\(Mg+2FeCl_3\rightarrow2FeCl_2+MgCl_2\)
a 2a 2a (mol)
\(Mg+FeCl_2\rightarrow Fe+MgCl_2\)
2a 2a 2a (mol)
$n_{Mg\ pư} = a + 2a = 3a(mol)$
$n_{Fe} = 2a(mol)$
$\Rightarrow m_{tăng} = m_{Fe} - m_{Mg\ pư} = 2a.56 - 3a.24 = 1,2$
$\Rightarrow a = 0,03$
$n_{FeCl_3} = 2a = 0,06(mol)$
$x = \dfrac{0,06}{0,25} = 0,24M$