\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2       0,4                      0,2  
\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\\ V_{H_2}=0,2.22,4=4,48\left(l\right)\\ pthh:FeO+H_2\underrightarrow{t^o}Fe+H_2O\) 
           0,2      0,2    0,2
\(m_{Fe}=0,2.56=11,2\left(g\right)\)

Ta có: \(n_{Mg}=\dfrac{14,4}{24}=0,6\left(mol\right)\)

a, PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

_____0,6____________________0,6 (mol)

\(\Rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\)

b, Ta có: \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

Xét tỉ lệ: \(\dfrac{0,2}{1}=\dfrac{0,6}{3}\), ta được pư hết.

Theo PT: \(n_{Fe}=2n_{Fe_2O_3}=0,4\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,4.56=22,4\left(g\right)\)

Bạn tham khảo nhé!

\(n_{Mg}=\dfrac{10,8}{24}=0,45\left(mol\right)\\ pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\) 
           0,45     0,45                          0,45 
\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\\ C\%_{H_2SO_4}=\dfrac{44,1}{176,4}.100\%=25\%\\ V_{H_2}=0,45.22,4=10,08\left(l\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\\ n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ n_{Zn}=n_{H_2SO_4}=n_{H_2}=0,05\left(mol\right)\\ m_{Zn}=0,05.65=3,25\left(g\right)\\ m_{\text{dd}H_2SO_4}=\dfrac{0,05.98}{19,6\%}=25\left(g\right)\\ V_{\text{dd}H_2SO_4}=\dfrac{25}{1,84}\approx13,587\left(ml\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

         0,2--->0,2--------->0,2------>0,2

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ m_{H_2SO_4}=\dfrac{0,2.98}{20\%}=98\left(g\right)\\ \rightarrow V_{ddH_2SO_4}=\dfrac{98}{1,14}=86\left(ml\right)=0,086\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,086}=2,33M\)

`Mg + H_2 SO_4 -> MgSO_4 + H_2`

`2/15`       `2/15`                               `2/15`          `(mol)`

`n_[Mg] = [ 3,2 ] / 24 = 2 / 15 (mol)`

`a)V_[H_2] = 2 / 15 . 22,4 ~~ 2,99(l)`

`b)m_[dd H_2 SO_4] = [ 2 / 15 . 98 ] / 20 . 100 ~~ 65,3(g)`

\(a,Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)=n_{H_2}=n_{H_2SO_4}\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,V_{ddH_2SO_4}=\dfrac{0,2}{2}=0,1\left(l\right)\)

a)

\(Mg + H_2SO_4 \to MgSO_4 + H_2\)

Magie tan dần, xuất hiện bọt khí không màu không mùi.

b)

\(n_{H_2} = n_{MgSO_4} = n_{Mg} = \dfrac{9,6}{24} = 0,4(mol)\\ m_{MgSO_4} = 0,4.120 = 48(gam)\\ V_{H_2} = 0,4.22,4 = 8,96(lít)\)

c)

\(CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Cu} = n_{H_2} = 0,4(mol)\\ \Rightarrow m_{Cu} = 0,4.64 = 25,6(gam)\)