a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\); \(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,8}{6}\) => Al hết, HCl dư

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

            0,2-->0,6---->0,2----->0,3

=> V_H2 = 0,3.22,4 = 6,72 (l)

b) \(\left\{{}\begin{matrix}m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\\m_{HCl\left(dư\right)}=\left(0,8-0,6\right).36,5=7,3\left(g\right)\end{matrix}\right.\)

=> m_{chất tan} = 26,7 + 7,3 = 34 (g)

c) m_{dd sau pư} = 5,4 + 200 - 0,3.2 = 204,8 (g)

\(\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{26,7}{204,8}.100\%=13,04\%\\C\%_{HCl\left(dư\right)}=\dfrac{7,3}{204,8}.100\%=3,56\%\end{matrix}\right.\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\) 
           0,2                       0,2       0,3 
\(V_{H_2}=0,3.22,4=6,72L\\ m_{AlCl_3}=133,5.0,2=26,7g\\ m_{\text{dd}}=5,4+200-\left(0,3.2\right)=204,8g\\ C\%=\dfrac{26,7}{204,8}.100\%=13\%\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{H_2SO_4}=0,5.0,1=0,05\left(mol\right)\\ pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) 
  \(LTL:\dfrac{0,2}{2}>\dfrac{0,05}{3}\) 
=> Al dư  
\(n_{H_2}=n_{H_2SO_4}=0,05\left(mol\right)\) 
\(V_{H_2}=0,05.22,4=1,12l\\ C_M=\dfrac{\dfrac{1}{60}}{0,1}=0,16M\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot29,4\%}{98}=0,6\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,6}{3}\) \(\Rightarrow\) Axit còn dư, Nhôm p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,6-0,3=0,3\left(mol\right)\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,3\cdot98=29,4\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72 \left(l\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Al}+m_{ddH_2SO_4}-m_{H_2}=204,8\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{204,8}\cdot100\%\approx16,7\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{29,4}{204,8}\cdot100\%\approx14,36\%\end{matrix}\right.\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

a)

PTHH: 

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2-->0,3------------------------->0,3

b) \(V_B=V_{H_2}=0,3.22,4=6,72\left(l\right)\)

c) \(C\%_{H_2SO_4}=\dfrac{0,3.98.100\%}{200}=14,7\%\)

a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,5.65=32,5\left(g\right)\)

\(\Rightarrow m_{CuO}=72,5-32,5=40\left(g\right)\)

c, Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Zn}+n_{CuO}=1\left(mol\right)\)

\(\Rightarrow b=C_{M_{H_2SO_4}}=\dfrac{1}{2,5}=0,4M\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnSO_4}=n_{Zn}=0,5\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnSO_4}}=\dfrac{0,5}{2,5}=0,2M\\C_{M_{CuSO_4}}=\dfrac{0,5}{2,5}=0,2M\end{matrix}\right.\)

Bạn tham khảo nhé!

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\left(I\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\left(II\right)\)

b, Theo PTHH(1) : \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{Zn}=32,5\left(g\right)\)

\(\Rightarrow m_{CuO}=m_{hh}-m_{Zn}=40\left(g\right)\)

\(\Rightarrow n_{CuO}=\dfrac{m}{M}=0,5\left(mol\right)\)

c, Theo PTHH (1) và (2) : \(n_{H2SO4}=n_{CuO}+n_{Zn}=1\left(mol\right)\)

\(\Rightarrow C_{MH2SO4}=b=\dfrac{n}{V}=\dfrac{1}{2,5}=0,4M\)

d, ( Chắc là thể tích coi như không đổi )

Thấy sau phản ứng thu được A gồm \(0,5molZnSO_4,0,5molCuSO_4\)

\(\Rightarrow C_{MCuSO4}=C_{MZnSO4}=\dfrac{n}{V}=\dfrac{0,5}{2,5}=0,2M\)

Vậy ...

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

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

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

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=11.2+200-0.2\cdot2=210.8\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{25.4}{210.8}\cdot100\%=12.05\%\)

Giúp với mai mình thi rồi!

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,3.24,79=7,437\left(l\right)\)

\(n_{Al}=\dfrac{5,4}{27}=0,2mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=1,5.0,2=0,3mol\\ V_{H_2}=0,3.24,79=7,437l\)

a)

$n_{Zn} = \dfrac{13}{65} = 0,2(mol) ; n_{H_2 SO_4} = 0,5.2 = 1(mol)$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
Ta thấy : 

$n_{Zn} < n_{H_2SO_4}$ nên $H_2SO_4$ dư

$n_{ZnSO_4} = n_{H_2SO_4\ pư} = n_{Zn} = 0,2(mol)$
$m_{ZnSO_4} = 0,2.161=32,2(gam)$
$m_{H_2SO_4\ pư} = 0,2.98 = 19,6(gam)$

b)

$n_{H_2SO_4\ dư} = 1 - 0,2 = 0,8(mol)$
$C_{M_{H_2SO_4\ dư}} = \dfrac{0,8}{0,5} = 1,6M$
$C_{M_{FeSO_4}} = \dfrac{0,2}{0,5} = 0,4M$

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{H_2SO_4}=0,5.2=1\left(mol\right)\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ a.Vì:\dfrac{0,2}{1}< \dfrac{1}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{Zn}=0,2\left(mol\right)\\ m_{H_2SO_4\left(p.ứ\right)}=0,2.98=19,6\left(g\right)\\ m_{ZnSO_4}=161.0,2=32,2\left(g\right)\\ b.V_{ddsau}=V_{ddH_2SO_4}=0,5\left(l\right)\\ C_{MddZnSO_4}=\dfrac{0,2}{0,5}=0,4\left(M\right)\\ C_{MddH_2SO_4\left(dư\right)}=\dfrac{1-0,2}{0,5}=1,6\left(M\right)\)