nH2=\(\dfrac{0,6}{2}=0,3\left(mol\right)\\\)
Gọi x, y lần lượt là số mol của Al , Al2O3
pthh:
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
x... ... ...3x... ... ...x... .... .... ..1,5x(mol)
Al2O3 + 6HCl \(\rightarrow\) 2AlCl3 + 3H2O
y... ... ... ... ..6y..... .....2y.. ... ... ..3y(mol)
Theo pthh:
nAl=x=\(\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\)
a, mAl=n.M=0,2.27=5,4(g)
\(\Rightarrow\) mAl2O3=25,8-5,4=20,4(g)\(\Rightarrow\) nAl2O3=y=\(\dfrac{20,4}{102}=0,2\left(g\right)\)
b, mdd AlCl3=mHCl+mhỗn hợp-mH2=200+25,8-0,6=225,2(g)
mAlCl3=(x+2y).M=(0,2+0,4).133,5=80,1(g)
\(\\\Rightarrow\) \(C\%_{ddAlCl_3}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{80,1}{225,2}.100\%=35,57\%\)
