\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

______0,05------>0,15--------->0,05

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)

\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)

PTHH: Fe₂(SO₄)₃ + 6NaOH --> 2Fe(OH)₃\(\downarrow\) + 3Na₂SO₄

________0,05----------------------->0,1

=> mFe(OH)₃ = 0,1.107=10,7(g)

\(m_{ct}=\dfrac{20.98}{100}=19,6\left(g\right)\)

\(n_{H2SO4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)

         1             1               1             1

        0,2          0,2           0,2

a) \(n_{CuO}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CuO}=0,2.80=16\left(g\right)\)

b) \(n_{CuSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CuSO4}=0,2.160=32\left(g\right)\)

c) \(m_{ddspu}=16+98=114\left(g\right)\)

\(C_{CuSO4}=\dfrac{32.100}{114}=28,7\)⁰/₀

 Chúc bạn học tốt

a) Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

b) \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH: Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

                0,1---->0,3------->0,1

=> m = 0,1.342 = 34,2 (g)

c) \(C\%_{dd.H_2SO_4}=\dfrac{0,3.98}{120}.100\%=24,5\%\)

Al2O3+3H2SO4->Al2(SO4)3+3H2O

0,1------0,3------0,1-------------------0,3

n Al2O3=0,1 mol

m Al2(SO4)3=0,1.342=34,2g

C%=\(\dfrac{0,3.98}{120}100=24,5\%\)

\(a.Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2 O\\ b.n_{Al_2O_3}=0,1\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,1\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\\ c.n_{H_2SO_4}=3n_{Al_2O_3}=0,3\left(mol\right)\\ C\%_{H_2SO_4}=\dfrac{0,3.98}{120}.100=24,5\%\)

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

\(m_{H_2SO_4}=200.19,6\%=39,2\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

a, PT: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

b, Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\), ta được H₂SO₄ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1\left(mol\right)\\n_{H_2SO_4\left(pư\right)}=3n_{Fe_2O_3}=0,3\left(mol\right)\end{matrix}\right.\)

⇒ n_{H2SO4 (dư)} = 0,4 - 0,3 = 0,1 (mol)

Ta có: m _{dd sau pư} = m_Fe2O3 + m _{dd H2SO4} = 16 + 200 = 216 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{216}.100\%\approx18,52\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1.98}{216}.100\%\approx4,54\%\end{matrix}\right.\)

Bạn tham khảo nhé!

\(n_{Fe_2O_3}=\dfrac{16}{160}=0.1\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{200\cdot19.6\%}{98}=0.4\left(mol\right)\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.4}{3}\Rightarrow H_2SO_4dư\)

\(m_{\text{dung dịch sau phản ứng}}=16+200=216\left(g\right)\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0.1\cdot400}{216}\cdot100\%=18.51\%\)

\(C\%_{H_2SO_4}=\dfrac{\left(0.4-0.2\right)\cdot98}{216}\cdot100\%=9.1\%\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right);n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\)

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2(SO_4)_3+3H_2O\)

Theo đề : 0,1.........0,4

Lập tỉ lệ : \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\)=> Sau phản ứng H2SO4 dư, Fe2O3 phản ứng hết

=> Dung dịch sau phản ứng gồm Fe2(SO4)3 và H2SO4 dư

m_{dd sau phản ứng} =16+200=216(g)

=>\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{216}=18,52\%\)

\(C\%_{H_2SO_4dư}=\dfrac{\left(0,4-0,1.3\right).98}{216}=0,45\%\)

\(n_{MgO}=a\left(mol\right),n_{Fe_2O_3}=b\left(mol\right)\)

\(m=40a+160b=12\left(g\right)\left(1\right)\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

\(m_{Muối}=120a+400y=32\left(g\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.05\)

\(\%MgO=\dfrac{0.1\cdot40}{12}\cdot100\%=33.33\%\)

\(\%Fe_2O_3=66.67\%\)

\(m_{dd}=12+200=212\left(g\right)\)

\(C\%_{MgSO_4}=\dfrac{0.1\cdot120}{212}\cdot100\%=5.66\%\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0.05\cdot400}{212}\cdot100\%=9.42\%\)

`Fe_2O_3+3H_2SO_4->Fe_2(SO_4)_3+3H_2O`

0,0625----------0,1875---------0,0625 mol

`->n_(Fe_2O_3)=10/160=0,0625mol`

`->m_(Fe_2(SO_4)_3)=0,0625.400=25g`

`->C%(H_2SO_4)=((0,1875.98)/(450)).100%=4,083%`

`#YBtran<3`

\(n_{Fe_2O_3}=\dfrac{10}{160}=0,0625\left(mol\right)\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,0625\left(mol\right)\\ a,m=m_{Fe_2\left(SO_4\right)_3}=400.0,0625=25\left(g\right)\\ b,n_{H_2SO_4}=3.0,0625=0,1875\left(mol\right)\\ C\%_{ddH_2SO_4}=\dfrac{0,1875.98}{450}.100\%\approx4,083\%\)