Question

Hoa tan hoan toan 0.2mol Fe va 0.1 mol Fe2O3 bang luong vua du dung dich H2SO4 dac dam thu duoc V lit SO2 (dkt?

hoa tan hoan toan 0.2mol Fe va 0.1 mol Fe2O3 bang luong vua du dung dich H2SO4 dac dam thu duoc V lit SO2 (dktc) va co can dung dich duoc m gam muoi khan

Answer 1

1.

2Fe + 6H₂SO_4(đ) \(\underrightarrow{t^o}\)Fe₂(SO₄)₃ + 3SO₂ + 6H₂O (1)

Fe + 3H₂SO_4(đ) \(\underrightarrow{t^o}\)Fe₂(SO₄)₃ + 3H₂O (2)

Theo PTHH 1 ta có:

\(\dfrac{3}{2}\)n_Fe=n_SO2=0,3(mol)

V_SO2=22,4.0,3=6,72(lít)

Answer 2

2.

2Fe + 6H₂SO_4(đ) \(\underrightarrow{t^o}\)Fe₂(SO₄)₃ + 3SO₂ + 6H₂O (1)

Fe₂O₃ + 3H₂SO_4(đ) \(\underrightarrow{t^o}\)Fe₂(SO₄)₃ + 3H₂O (2)

Theo PTHH 1 và 2 ta có:

\(\dfrac{1}{2}\)n_Fe=n_Fe2(SO4)3=0,1(mol)

n_Fe2O3=n_Fe2(SO4)3=0,1(mol)

m_muối=0,2.400=80(g)

Answer 3

2.nFe(Fe2O3)=0.1.2=0.2mol
nFe trc pứ=0.2+0.2=0.4mol
nFe[Fe2(SO4)3]=nFe trc pứ=0.4mol
=>nFe(SO4)3=0.4/2=0.2mol
mFe2(SO4)3=400.0.2=80g.
Vậy m=80g.

Answer 4

nFe(Fe2O3)=0.1.2=0.2mol
nFe trc pứ=0.2+0.2=0.4mol
nFe[Fe2(SO4)3]=nFe trc pứ=0.4mol
=>nFe(SO4)3=0.4/2=0.2mol
mFe2(SO4)3=400.0.2=80g.
Vậy m=80g.