Question

hoà tan 27,2 g hỗn hop gồm Fe₂o₃ va Zn trong 500 ml dd HCl vừa đủ thấy thoát ra4,48 lít H2 ở dktc .

A) Tính C% khôi luong cac chat ban dau B) Tính m_muoi

Answer 1

a)\(n_{H_2}\)=4,48:22,4=0,2(mol)

Ta có PTHH:

Fe₂O₃+6HCl->2FeCl₃+3H₂O(1)

0,08875............0.1775..................(mol)

Zn+2HCl->ZnCl₂+H₂(2)

0,2.................0,2.......0,2...(mol)

Theo PTHH(2):n_Zn=0,2(mol)

=>m_Zn=0,2.65=13(g)

Mặt khác: m_hhKL=\(m_{Fe_2O_3}\)+m_Zn=27,2

=>\(m_{Fe_2O_3}\)=27,2-m_Zn=27,2-13=14,2(g)

=>\(n_{Fe_2O_3}\)=14,2:160=0,08875(mol)

Vậy C_%Zn=\(\dfrac{13}{27,2}\).100%=47,8%

\(C_{\%Fe_2O_3}\)=\(\dfrac{14,2}{27,2}\).100%=52,2%

b)Theo PTHH(1);(2):\(m_{FeCl_3}\)=0,1775.162,5=28,84375(g)

\(m_{ZnCl_2}\)=0,2.136=27,2(g)

Vậy m_muối=\(m_{FeCl_3}\)+\(m_{ZnCl_2}\)=28,84375+27,2=56,04375(g)