Kiểm tra 1 tiết Hiđro: Đề 2

nCu=50,4/64=0,7875(mol)

CuO+H2--->Cu+H2O

0,7875_____0,7875

mCuO=0,7875.80=63(g)

===>mCuO cần dùng =63.10/80=78,75(g)

gọi số mol 2 KL Al và Zn đều (là a mol)

2Al + 6HCl => 2AlCl₃ + 3H₂

a---------------------------> 1,5a

Zn + 2HCl => ZnCl₂ + H₂

a-------------------------------> a

ta có: n_H2 = 1,5a +a = 2,5 a = 0,15 mol

=> a= 0,06 mol <=> n_Al = n_Zn = 0,06 mol

b) m_{Zn = 0,06 * 65 = 3,9 (g)}

mAl = 0,06 * 27 = 1,2 (g)

pt:

2Al+6HCl--->2AlCl3+3H2

a____3a______a____3/2a

Zn+2HCl--->ZnCl2+H2

a____2a_____a____a

Ta có: 3/2a+a=0,15=>a=0,06

=>nl=nZn=0,06(mol)

b)mAl=0,06.27=1,62(g)

mZn=0,06.65=3,9(g)

nCuO=32/80=0,4(mol)

CuO+H2-t*-->Cu+H2O

0,4__0,4___0,4

=>VH2=0,4.22,4=8,96(l)

mCu=0,4.64=25,6(g)

2Cu+O2-t*-->2CuO

2____1

0,4___0,4

Ta có; 0,4/2<0,4/1

=>O2 dư

Theo pt: nCuO=nCu=0,4(mol)

=>mCuO=0,4.80=32(g)

Nhóm 1:

n_HCl = \(\dfrac{7,3}{36,5}=0,2\) mol

Pt: 2Al + 6HCl --> 2AlCl₃ + 3H₂

................0,2 mol------------> 0,1 mol

V_H2 = 0,1 . 22,4 = 2,24 (lít)

Nhóm 2:

n_H2SO4 = \(\dfrac{9,8}{98}=0,1\) mol

Pt: Zn + H₂SO₄ --> ZnSO₄ + H₂

...............0,1 mol--------------> 0,1 mol

V_H2 = 0,1 . 22,4 = 2,24 (lít)

Dẫn toàn bộ lượng khí H2 qua đâu bạn?

a) \(pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\left(1\right)\)

\(\text{b) }n_{Zn}=\dfrac{m}{M}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theo \(pthh\left(1\right):n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow m_{ZnSO_4}=n\cdot M=0,1\cdot161=16,1\left(g\right)\)

c) \(n_{H_2}=\dfrac{v}{22,4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Theo\text{ }pthh:1mol\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }1mol\\ Theo\text{ }đb:\text{ }0,1mol\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0,05\left(mol\right)\\ Pứ:\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{0,05mol}{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{0,05mol}{ }\\ Sau\text{ }pứ:\text{ }\text{ }\text{ }\text{ }0,05mol\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0\)

\(\Rightarrow\) Sau phản ứng: \(Zn\) còn dư

- Rượu etylic và nước

- Bezen và dầu ăn

- muối và nước

- 1 ml dầu và 2ml xăng

- khí amoniac ( NH3) và nước