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

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

a, \(n_{H_2SO_4}=3n_{Fe_2O_3}=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4}{9,8\%}=300\left(g\right)\)

b, \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 16 + 300 = 316 (g)

\(\Rightarrow C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{316}.100\%\approx12,66\%\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

a, \(V_{H_2}=0,1.24,79=2,479\left(l\right)\)

b, \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{9,8}{9,8\%}=100\left(g\right)\)

c, Ta có: m _{dd sau pư} = 5,6 + 100 - 0,1.2 = 105,4 (g)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,1.152}{105,4}.100\%\approx14,42\%\)

1. \(Fe+H_2SO_{4\left(đ\right)}\underrightarrow{t^o}Fe_2\left(SO_4\right)_3+H_2S+H_2O\)

\(2Fe^0\rightarrow2Fe^{+3}+6e|\times4\)

\(S^{+6}+8e\rightarrow S^{-2}|\times3\)

→ \(8Fe+15H_2SO_{4\left(đ\right)}\underrightarrow{t^o}4Fe_2\left(SO_4\right)_3+3H_2S+12H_2O\)

2. \(HNO_3+H_2S\rightarrow NO+S+H_2O\)

\(N^{+5}+3e\rightarrow N^{+2}|\times2\)

\(S^{-2}\rightarrow S^0+2e|\times3\)

→ \(2HNO_3+3H_2S\rightarrow2NO+3S+4H_2O\)

a, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Ta có: \(n_{KCl}=\dfrac{1,49}{74,5}=0,02\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KCl}=0,03\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,03.24,79=0,7437\left(l\right)\)

b, Theo PT: \(n_{KClO_3\left(TT\right)}=n_{KCl}=0,02\left(mol\right)\)

\(\Rightarrow m_{KClO_3\left(TT\right)}=0,02.122,5=2,45\left(g\right)\)

\(\Rightarrow H=\dfrac{2,45}{3,5}.100\%=70\%\)

2KClO3=>2KCl+3O2

a, nKCl=1,49/74,5=0,02(mol)

=>nO2=0,03(mol)

=>V O2=0,03.22,4=0,672(l)

b, nKClO3=0,02(mol)

mKClO3=0,02.122,5=2,45(g)

H(KClO3)=2,45/3,5.100%=70%

cảm ơn trước nha

1 mol CuO tác dụng với H₂SO₄ thu được 1 mol CuSO₄ 

a, \(n_{Al_2O_3}=\dfrac{15,3}{102}=0,15\left(mol\right)\)

\(m_{H_2SO_4}=300.19,6\%=58,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)

PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,6}{3}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=3n_{Al_2O_3}=0,45\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,6-0,45=0,15\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,15.98=14,7\left(g\right)\)

b, Có lẽ đề hỏi khối lượng các chất sau pư chứ nhỉ?

Theo PT: \(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,15\left(mol\right)\\n_{H_2O}=3n_{Al_2O_3}=0,45\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)

\(m_{H_2O}=0,45.18=8,1\left(g\right)\)

m_{H2SO4 (dư)} = 14,7 (g)

a, \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

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

b, \(n_{C_2H_4}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

\(n_{CO_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)