2Al+3H2SO4->Al2(SO4)3+3H2

0,1----------------------0,075----0,15

n H2=0,15 mol

=>mAl=0,1.27=2,7g

=>m Al2(SO4)3=0,075.342=25,65g

a) PTHH: \(2Al+3H_2SO_2\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{Al}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)

\(m_{Al}=0,1.27=2,7\left(g\right)\)

c) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

2Al+3H2SO4->Al2(SO4)3+3H2

0,1----------------------0,075----0,15

n H2=0,15 mol

=>mAl=0,1.27=2,7g

=>m Al2(SO4)3=0,075.342=25,65g

Ta có: \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)

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

______0,5____0,5_____0,5_____0,5 (mol)

b, m_H2SO4 = 0,5.98 = 49 (g)

c, m_FeSO4 = 0,5.152 = 76 (g)

d, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

____0,5__0,5 (mol)

⇒ m_Cu = 0,5.64 = 32 (g)

Bạn tham khảo nhé!

Sửa đề : 11.2 g sắt

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

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

\(0.2....0.2.................0.2\)

\(m_{FeSO_4}=0.2\cdot152=30.4\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{0.2}{0.05}=4\left(M\right)\)

\(m_{Fe}=\dfrac{11}{56}=0.19\left(mol\right)\)

\(Fe+H_2SO_4->FeSO_4+H_2\)

0.19   0.19            0.19              (mol)

\(m_{FeSO_4}=0.19\cdot152=28.28\left(g\right)\)

\(Cm=\dfrac{0.19}{0.05}=3.8M\)

nFe=11.256=0.2(mol)

mFeSO4=0,2⋅152=30,4(g)

`a)PTHH:`

`Fe+H_2 SO_4 ->FeSO_4 +H_2 \uparrow`

`0,2`     `0,2`             `0,2`           `0,2`                     `(mol)`

`b)n_[Fe]=[11,2]/56=0,2(mol)`

   `n_[H_2 SO_4]=[294.10]/[100.98]=0,3(mol)`

Ta có: `0,2 < 0,3=>Fe` hết, `H_2 SO_4` dư

`@m_[FeSO_4]=0,2.152=30,4(g)`

`@V_[H_2]=0,2.22,4=4,48(l)`

`@C%_[FeSO_4]=[30,4]/[11,2+294-0,2.2].100=9,97%`

$n_{Fe}=\frac{19,6}{56}=0,35(mol)$

$Fe+H_2SO_4\to FeSO_4+H_2$

Theo PT: $n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=0,35(mol)$

$\to m_{H_2SO_4}=0,35.98=34,3(g)$

$m_{FeSO_4}=0,35.152=53,2(g)$

$m_{H_2}=0,35.2=0,7(g)$

\(BTKL:m_{Fe}+m_{H_2O_4}=m_{FeSO_4}+m_{H_2}\\ \Rightarrow m_{H_2}=11,2+19,6-30,4=0,4(g)\)

a) nFe=0,1(mol); nHCl=0,4(mol)

PTHH: Fe + 2 HCl -> FeCl2 + H2

Ta có: 0,1/1 < 0,4/2

=> Fe hết, HCl dư, tish theo nFe.

b) nH2=nFeCl2=Fe=0,1(mol)

=> V(H2,đktc)=0,1.22,4=2,24(l)

c) mFeCl2=127.0,1=12,7(g)

a) nFe=0,1(mol); nHCl=0,4(mol) PTHH: Fe + 2 HCl -> FeCl2 + H2 Ta có: 0,1/1 < 0,4/2 => Fe hết, HCl dư, tish theo nFe. b) nH2=nFeCl2=Fe=0,1(mol) => V(H2,đktc)=0,1.22,4=2,24(l) c) mFeCl2=127.0,1=12,7(g)

Gọi \(n_{Fe}=x\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{FeCl_2}=n_{Fe}=x\left(mol\right)\)

Vì khối lượng muối FeCl2 tăng 7,1g so với khối lượng bột Fe

\(\Rightarrow127x-56x=7,1\\ \Rightarrow x=0,1\)

\(n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ V_{H_2\left(ĐKTC\right)}=0,1.22,4=2,24\left(l\right)\)

Chọn D

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(n_{FeSO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeSO_4}=0,2.152=30,4\left(g\right)\)

c, \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

d, \(n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)