\(n_{Fe_2O_3}=\dfrac{4}{160}=0,025\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O

Mol:     0,025        0,075           0,025

\(m_{ddH_2SO_4}=\dfrac{0,075.98.100}{9,8}=75\left(g\right)\)

m_{dd sau pứ} = 4 + 75 = 79 (g)

\(C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,025.400.100\%}{79}=12,66\%\)

giúp mình với ạ

Fe2O3 +3H2SO4----.Fe2(SO4)3 +3H2O

a) Ta có

n\(_{Fe2O3}=\frac{4}{160}=0,025\left(mol\right)\)

Theo pthh

n\(_{H2SO4}=3n_{Fe}=0,075\left(mol\right)\)

m\(_{H2SO4}=0,075.98=7,35\left(g\right)\)

b)m\(_{ddH2SO4}=\frac{7,35.100}{9,8}=75\left(g\right)\)

c) Theo pthh

n\(_{Fe2\left(SO4\right)3}=n_{Fe}=0,025\left(mol\right)\)

m\(_{Fe2\left(SO4\right)3}=0,025.400=10\left(g\right)\)

C%=\(\frac{10}{75+4}=12,66\%\)

Chúc bạn học tốt

PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

a+b) Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{FeCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,6\cdot36,5}{14,6\%}=150\left(g\right)\\m_{FeCl_3}=0,2\cdot162,5=32,5\left(g\right)\end{matrix}\right.\)

\(\Rightarrow C\%_{FeCl_3}=\dfrac{32,5}{150+16}\cdot100\%\approx19,58\%\)

b) PTHH: \(HCl+KOH\rightarrow KCl+H_2O\)

Theo PTHH: \(n_{KOH}=n_{HCl}=0,6\left(mol\right)\) \(\Rightarrow V_{KOH}=\dfrac{0,6}{0,5}=1,2\left(l\right)\)

\(M+H_2SO_4\rightarrow MSO_4+H_2\uparrow\\ n_{ASO_4}=n_A=n_{H_2}=n_{H_2SO_4}=a\left(mol\right)\\ 1.m_{ddH_2SO_4}=\dfrac{98a.100}{20}=490a\left(g\right)\\ 2.m_{ddsau}=M_M.a+490a-2a=\left(M_M+488\right).a\left(g\right)\\ C\%_{ddsau}=22,64\%\\ \Leftrightarrow\dfrac{\left(M_M+96\right)a}{\left(M_M+488\right)a}.100\%=22,64\%\\ \Leftrightarrow M_M=18,72\left(loại\right)\)

Khả năng cao sai đề nhưng làm tốt a,b nha

n_Fe2O3 = \(\dfrac{4}{160}\) = 0,025 mol

Fe₂O₃ + 3H₂SO₄ -> Fe₂(SO₄)₃ + 3H₂O

0,025->0,075------->0,025 mol

=>m_d2(H2SO4) = \(\dfrac{0,075.98.100}{9,8}\) = 75 g

C%_Fe2(SO4)3 = \(\dfrac{0,025.400}{4+75}.100\%\) = 12,65%