Gọi nFe= x (mol); nMg= y(mol)
PTHH: Fe +2 HCl \(\rightarrow\) FeCl2 + H2\(\uparrow\)
x-> 2x x x (mol)
Mg + 2HCl \(\rightarrow\) MgCl2 + H2\(\uparrow\)
y-> 2y y y (mol)
Ta có: mdd sau p.ứng = \(56x+24y+\frac{\left(73x+73y\right).100}{20}-\left(2x+2y\right)\)
= 419x + 387y (g)
mFeCl2 = 127x (g)
\(\Rightarrow\) C%FeCl2 = \(\frac{127x}{419x+387y}.100\) = 15,76%
\(\Leftrightarrow\) \(\frac{127x}{419x+387y}\) = 0,1576
\(\Leftrightarrow\) \(\frac{x}{y}=1\) \(\Leftrightarrow\) x=y
Ta có: mMgCl2 = 95y = 95x(g)
m dd sau p.ứng = 419x + 387y = 419x + 387x = 806x (g)
\(\Rightarrow\) C% MgCl2 = \(\frac{95x}{806x}.100\) = 11,79%
