x - nr. moli FeCl2
y - nr moli FeCl3
M FeCl2 = 56+2*35,5 = 127 g/mol
M FeCl3 = 56+3*35,5 = 162,5 g/mol
m amestec = m FeCl2 + m FeCl3 = x*127+y*162,5
%Cl = m Cl/m amestec * 100
60,35 = (71x+106,5y)*100/(127x+162,5y)
7100+10650y = 7664,45x + 9806,875y
843,125y = 564,45x 8*6 = 24*2
2/6 =
x/y = 564,54/843,125 = 0,66 => x/y = 2/3