Tóm tắt

\(V_{H_2\left(đktc\right)}=8,96l\\ C_{\%H_2SO_4}=19,6\%\\ a)m_{Zn}=?\\ m_{ddH_2SO_4}=?\\ b)C_{\%ZnSO_4}=?\)

\(a)n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

0,4        0,4              0,4           0,4

\(m_{Zn}=0,4.65=26g\\ m_{ddH_2SO_4}=\dfrac{0,4.98}{19,6}\cdot100=200g\\ b)C_{\%ZnSO_4}=\dfrac{0,4.161}{26+200-0,4.2}\cdot100=28,6\%\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH :

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

 0,4    0,4             0,4        0,4

\(a,m_{Zn}=0,4.65=26\left(g\right)\)

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

\(m_{ddH_2SO_4}=\dfrac{39,2.100}{19,6}=200\left(g\right)\)

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

\(m_{ddZnSO_4}=200+26-\left(0,4.2\right)=225,2\left(g\right)\)

\(C\%_{ZnSO_4}=\dfrac{64,4}{225,2}.100\%\simeq28,6\%\)

a) 

\(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)

PTHH: Zn + H₂SO₄ --> ZnSO₄ + H₂

_____0,4<---0,4<--------0,4<----0,4

=> m_Zn = 0,4.65 = 26 (g)

=> \(\left\{{}\begin{matrix}\%Zn=\dfrac{26}{51,6}.100\%=50,388\%\\\%Cu=\dfrac{51,6-26}{51,6}.100\%=49,612\text{%}\end{matrix}\right.\)

b) 

m_ZnSO4 = 0,4.161 = 64,4 (g)

c) 

\(V_{ddH_2SO_4}=\dfrac{0,4}{2}=0,2\left(l\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

Ta có: \(n_{H_2SO_4}=0,3\cdot1=0,3\left(mol\right)=n_{Zn}=n_{ZnSO_4}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,3\cdot65=19,5\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\C_{M_{ZnSO_4}}=\dfrac{0,3}{0,3}=1\left(M\right)\end{matrix}\right.\)

\(Mg + H_2SO_4 \to MgSO_4 + H_2\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ Zn + H_2SO_4 \to ZnSO_4 + H_2\\ n_{H_2} = n_{H_2SO_4} = a(mol)\\ \text{Bảo toàn khối lượng : }\\ 20 + 98a = 58,4 + 2a\\ \Rightarrow a = 0,4\\ \Rightarrow V_1 = \dfrac{0,4}{0,5} = 0,8(lít) = 800(ml)\\ V_2 = 0,4.22,4 = 8,96(lít)\)

a, PTPƯ: SO₃ + H₂O ---> H₂SO₄

n_SO3=\(\dfrac{2,24}{22,4}=0,1mol\)

1 mol SO₃ ---> 0,1 mol H₂SO₄

nên 0,1 mol SO₃ ---> 0,1 mol H₂SO₄

C_{M H2SO4}=\(\dfrac{0,1}{0,5}\)=0,2 M

b, PTPƯ: Zn + H₂SO₄ ---> ZnSO₄ + H₂

1 mol H₂SO₄ ---> 1 mol Zn

nên 0,1 mol H₂SO₄ ---> 0,1 mol Zn

m_Zn=0,1.65=6,5 g

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

Ta có: \(n_{Zn}=\dfrac{4,8}{65}=\dfrac{24}{325}\left(mol\right)\)

Đến đây thì ra số mol hơi xấu, bạn xem lại đề nhé.

\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{H_2SO_4}=n_{ZnSO_4}=n_{H_2}=0,1mol\\ C_{M_{H_2SO_4}}=\dfrac{0,1}{0,2}=0,5M\\ b)m_{ZnSO_4}=0,1.161=16,1g\)

n_H2SO4 = 49/98 = 0.5 (mol) 

C_MH2SO4 = 0.5/0.15 = 3.3 (M) 

Zn + H₂SO₄ => ZnSO₄  + H₂

...........0.5.............0.5.........0.5

V_H2 = 0.5 * 22.4 = 11.2 (l) 

C_MZnSO4 = 0.5 / 0.15 = 10/3 (M) 

C%_ZnSO4 = C_M*M / 10D = 10/3 * 161 / 10 * 1.25 = 42.9 %