Thiếu H2SO4 tham gia pư là 49 g!
Oxit kim loại R2Ox
\(n_{H2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
\(R_2O_x+xH_2\rightarrow2R+xH_2O\)
0,4/x____0,4mol_____0,8/x mol
\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{H2SO4}=\frac{49}{98}=0,5\left(mol\right)\)
\(R+yH_2SO_4\rightarrow R_2\left(SO_4\right)_y+yH_2\)
________ 0,3__________0,3 _________
\(R_2O_x+xH_2SO_4\rightarrow R_2\left(SO_4\right)_x+xH_2O\)
0,2/x_______0,2_____________________
\(n_{R2Ox}=\left(\frac{0,4}{x}+\frac{0,2}{x}\right).\left(2R+16x\right)=34,8\)
Biện luận :
\(x=1\Rightarrow R=21\left(loai\right)\)
\(x=2\Rightarrow R=42\left(laoi\right)\)
\(x=3\Rightarrow R=56\left(Fe\right)\)
Vậy CTHH là Fe2O3
