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

\(a,PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ b,n_{FeCl_2}=n_{H_2}=n_{Fe}=0,4\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ c,m_{FeCl_2}=127.0,4=50,8\left(g\right)\)

Fe+H2SO4->FeSO4+H2

0,1----0,1-------0,1-----0,1 

n Fe=5,6\56=0,1 mol

=>VH2=0,1.22,4=2,24l

=>m H2SO4=0,1.98=9,8g

=>C%=9,8\100 .100=9,8%

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

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

\(0.25....0.25.......................0.25\)

\(m_{Fe}=0.25\cdot56=14\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{0.25}{0.1}=2.5\left(M\right)\)

a. PTHH: Fe+H2SO4-->FeSO4+H2

b. Có nH2=\(\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

Theo pthh trên, nFe=nH2=0.25mol

=> mFe=0.25*56=14g

c. Theo pthh trên, nH2=nH2SO4=0.25mol

Đổi 100ml=0.1l

=> \(C_M=\dfrac{0.25}{0.1}=2.5M\)

a)

$Fe + CuSO_4 \to FeSO_4 + Cu$

b) Theo PTHH :

n Fe = n Cu = n FeSO4 = n CuSO4 = 200.16%/160 = 0,2(mol)

m Fe = 0,2.56 = 11,2(gam)

c) m Cu = 0,2.64 = 12,8(gam)

d) m dd = 11,2 + 200 -12,8 = 198,4(gam)

m FeSO4 = 0,2.152 = 30,4(gam)

C% FeSO4 = 30,4/198,4   .100% = 15,32%

Ta có: m_CuSO4 = 200.16% = 32 (g)

\(\Rightarrow n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)

a, PT: \(Fe+CuSO_4\rightarrow FeSO_4+Cu_{\downarrow}\)

_____0,2____0,2______0,2_____0,2 (mol)

b, m_Fe = 0,2.56 = 11,2 (g)

c, m_Cu = 0,2.64 = 12,8 (g)

d, Ta có: m _{dd sau pư} = m_Fe + m _{dd CuSO4} - m_Cu = 11,2 + 200 - 12,8 = 198,4 (g)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,2.152}{198,4}.100\%\approx15,32\%\)

Bạn tham khảo nhé!

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

PT: \(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)

a, Theo PT: \(n_{CH_3COOH}=2n_{Fe}=0,2\left(mol\right)\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{10\%}=120\left(g\right)\)

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

b, Theo PT: \(n_{\left(CH_3COO\right)_2Fe}=n_{Fe}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 5,6 + 120 - 0,1.2 = 125,4 (g)

\(\Rightarrow C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{0,1.174}{125,4}.100\%\approx13,88\%\)

\(n_{HCl}=0,25.2=0,5\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{Zn}=0,25.65=16,25\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

Không biết đúng không nữa;-;;;

a) PTHH: Zn + 2HCl -> ZnCl2 + H2 

b) HCl=250ml=0,25l

n2HCl= V/22,4= 0,5/22,4= 0,02(mol)

Zn  +  2HCl -> ZnCl2  +  H2

1          2              1         1

0,01 <-0,5--------------> 0,01

mZn= n.M= 0,01.65= 0,65(gam)

c) VH2=n . 22,4= 0,01 . 22,4= 0,224(l)
 

`Fe+2HCl->FeCl_2+H_2`

0,15--0,3---0,15----0,15 mol

`n_(Fe)=(8,4)/56=0,15 mol`

`->V_(H_2)=0,15.22,4=3,36l`

c) `CuO+H_2->Cu+H_2O`(to)              

             0,15---0,15 mol

`n_(CuO)=16/80=0,2 mol`

=>CuO dư

`->m_(Cu)=0,15.64=9,6g`

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\)

Phản ứng thế

\(b,n_{Zn}=\dfrac{1,3}{65}=0,02(mol)\\ \Rightarrow n_{ZnCl_2}=n_{H_2}=0,02(mol)\\ \Rightarrow m_{ZnCl_2}=0,02.136=2,72(g)\\ V_{H_2}=0,02.22,4=0,448(l)\)