a) Zn + H₂SO₄ --> ZnSO₄ + H₂

\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)

PTHH: Zn + H₂SO₄ --> ZnSO₄ + H₂

            0,4-->0,4------->0,4---->0,4

=> \(m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)

b) \(m_{ZnSO_4}=0,4.161=64,4\left(g\right)\)

c) \(\left\{{}\begin{matrix}m_{H_2}=0,4.2=0,8\left(g\right)\\V_{H_2}=0,4.22,4=8,96\left(l\right)\end{matrix}\right.\)

\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\) 
           0,4                      0,4           0,4 
\(m_{H_2SO_4}=0,8.98=78,4\left(g\right)\\ m_{ZnSO_4}=136.0,4=54,4\left(g\right)\\ m_{H_2}=0,4.2=0,6\left(g\right)\\ V_{H_2}=0,4.22,4=8,96\left(l\right)\)

Ta có: \(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)

a, PT: \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

______0,2_____0,4____0,2 (mol)

b, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{14,6}{10\%}=146\left(g\right)\)

c, \(C\%_{ZnCl_2}=\dfrac{0,2.136}{16,2+146}.100\%\approx16,77\%\)

Bạn tham khảo nhé!

`Fe + H_2 SO_4 -> FeSO_4 + H_2`

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

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

`a) V_[H_2] = 0,2 . 22,4 = 4,48 (l)`

`b) m_[dd HCl] = [ 0,2 . 98 ] / [ 9,8 ] . 100 = 200 (g)`

`c) C%_[FeSO_4] = [ 0,2 . 152 ] / [ 11,2 + 200 - 0,2 . 2 ] . 100 ~~ 14,42%`

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=0,1\left(mol\right);n_{HCl}=2.0,1=0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ b,m_{HCl}=0,2.36,5=7,3\left(g\right)\\ m_{FeCl_2}=127.0,1=12,7\left(g\right)\)

Cái khí ở dạng phân tử nên là H₂ chứ không phải H em nha!

\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

______0,05------>0,15--------->0,05

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)

\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)

PTHH: Fe₂(SO₄)₃ + 6NaOH --> 2Fe(OH)₃\(\downarrow\) + 3Na₂SO₄

________0,05----------------------->0,1

=> mFe(OH)₃ = 0,1.107=10,7(g)

$a\big)$

$Zn+2HCl\to ZnCl_2+H_2$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$b\big)$

$n_{Zn}=\dfrac{10,4}{65}=0,16(mol)$

Theo PT: $n_{Cu}=n_{Zn}=0,16(mol)$

$\to m_{Cu}=0,16.64=10,24(g)$

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

Hiện tượng: viên kẽm tan dần, có khí không màu thoát ra.

\(b,n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\\ \Rightarrow n_{HCl}=2n_{Zn}=0,3\left(mol\right)\\ \Rightarrow m_{HCl}=0,3\cdot36,5=5,475\left(g\right)\\ c,n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,15\cdot136=20,4\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\\ \Rightarrow m_{dd_{ZnCl_2}}=9,75+100-0,3=109,45\left(g\right)\\ \Rightarrow C\%_{dd_{ZnCl_2}}=\dfrac{20,4}{109,45}\cdot100\%\approx18,64\%\)