\(n_{Fe}=\frac{2,24}{5,6}=0,04\left(mol\right)\)
\(n_{Al}=\frac{m}{27}\left(mol\right)\)
PTHH: Fe+2HCl→ FeCl2+ H2 ( ↑ ) ( 1)
2Al+ 3H2SO4 →Al2(SO4)3 + 3H2 ( ↑ ) (2)
Theo PTHH (1) → nH2 (1)
= nFe = 0,04 ( mol ) → mH2 (1)= 0,04 . 2 = 0,08 ( g )
Theo PTHH (2) → nH2 ( 2) = \(\frac{3}{2}\) .nAl = \(\frac{3}{2}\) .\(\frac{m}{27}\)= \(\frac{m}{18}\)\(\rightarrow\) mH2(2) = \(\frac{m}{18}.2=\frac{m}{9}\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}\Delta mA=mFe-mH2\left(1\right)=2,24-0,08=2,16\left(g\right)\\\Delta mB=mAl-mH2\left(2\right)=m=\frac{m}{9}=\frac{8m}{9}\left(g\right)\end{matrix}\right.\)
Mà cân ở vị trí cân bằng \(\rightarrow\Delta mA=\Delta mB\)
\(\rightarrow2,16=\frac{8m}{9}\rightarrow m=2,43\left(g\right)\)