Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(n_{Fe_2O_3}=\dfrac{4}{160}=0,025\left(mol\right)\)
a. PTHH: Fe2O3 + 3H2SO4 ---> Fe2(SO4)3 + 3H2O
Theo PT: \(n_{H_2SO_4}=3.n_{Fe_2O_3}=3.0,025=0,075\left(mol\right)\)
=> \(m_{H_2SO_4}=0,075.98=7,35\left(g\right)\)
b. Ta có: 
\(C_{\%_{H_2SO_4}}=\dfrac{7,35}{m_{dd_{H_2SO_4}}}.100\%=9,8\%\)
=> \(m_{dd_{H_2SO_4}}=75\left(g\right)\)
Ta có: \(m_{dd_{Fe_2\left(SO_4\right)_3}}=75+4=79\left(g\right)\)
Theo PT: \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,025\left(mol\right)\)
=> \(m_{Fe_2\left(SO_4\right)_3}=0,025.400=10\left(g\right)\)
=> \(C_{\%_{Fe_2\left(SO_4\right)_3}}=\dfrac{10}{79}.100\%=12,66\%\)
Ta có: \(n_{Fe_2O_3}=\dfrac{9,6}{160}=0,06\left(mol\right)\)
a. PTHH: Fe2O3 + 3H2SO4 ---> Fe2(SO4)3 + 3H2O (1)
Theo PT(1): \(n_{H_2SO_4}=3.n_{Fe_2O_3}=3.0,06=0,18\left(mol\right)\)
=> \(m_{H_2SO_4}=0,18.98=17,64\left(g\right)\)
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{17,64}{m_{dd_{H_2SO_4}}}.100\%=9,8\%\)
=> \(m_{dd_{H_2SO_4}}=180\left(g\right)\)
b. Ta có: \(m_{dd_{Fe_2\left(SO_4\right)_3}}=9,6+180=189,6\left(g\right)\)
Theo PT(1): \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,06\left(mol\right)\)
=> \(m_{Fe_2\left(SO_4\right)_3}=0,06.400=24\left(g\right)\)
=> \(C_{\%_{Fe_2\left(SO_4\right)_3}}=\dfrac{24}{189,6}.100\%=12,66\%\)
c. PTHH: Fe2(SO4)3 + 3BaCl2 ---> 3BaSO4↓ + 2FeCl3 (2)
Theo PT(2): \(n_{BaSO_4}=3.n_{Fe_2\left(SO_4\right)_3}=3.0,06=0,18\left(mol\right)\)
=> \(m_{BaSO_4}=0,18.233=41,94\left(g\right)\)
Theo PT(2): \(n_{BaCl_2}=n_{BaSO_4}=0,18\left(mol\right)\)
=> \(m_{BaCl_2}=0,18.208=37,44\left(g\right)\)
Ta có: \(C_{\%_{BaCl_2}}=\dfrac{37,44}{m_{dd_{BaCl_2}}}.100\%=10,4\%\)
=> \(m_{dd_{BaCl_2}}=360\left(g\right)\)
\(n_{Al}=\dfrac{10,8}{27}=0,4(mol)\\ a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,6(mol);n_{Al_2(SO_4)_3}=0,2(mol)\\ b,V_{H_2}=0,6.22,4=13,44(l)\\ c,m_{dd_{H_2SO_4}}=\dfrac{0,6.98}{9,8\%}=600(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,2.342}{10,8+600-0,6.2}.100\%=11,22\%\)
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
\(n_{Fe_2O_3}=\dfrac{4}{160}=0,025\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O
Mol: 0,025 0,075 0,025
\(m_{ddH_2SO_4}=\dfrac{0,075.98.100}{9,8}=75\left(g\right)\)
mdd 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