a) Fe2O3+ 6HCl---->2FeCl3 +3H2O
0,1-----------0,6---------0,2-------0,3
b) Ta có
n\(_{Fe2O3}=\frac{16}{160}=0,1\left(mol\right)\)
m\(_{HCl}=\frac{365.20}{100}=73\left(g\right)\)
n\(_{HCl}=\frac{73}{36,5}=2\left(mol\right)\)
=> HCl dư
Theo pthh
Khí đâu ra nhỉ?
c) Theo pthh
n\(_{FeCl3}=2n_{Fe2O3}=0,2\left(mol\right)\)
m\(_{FeCl3}=0,2.162,5=32,5\left(g\right)\)
d) n\(_{HCl}dư=1,2\left(g\right)\)
C% HCl =\(\frac{1,4.36,5}{365+16}.100\%=13,41\%\)
C% FeCl3 =\(\frac{0,2.162,5}{365+16}.100\%=8,53\%\)
n