Question

dot nong 1,35(g) bot nhom trong khi clo du sau phan ung thu duoc nhom clorua (ALCL ₃)

a, lap phuong trinh phan ung hoa hoc xay ra

b, tinh khoi luong cua nhom clorua thu duocva the tich khi clo tham gia (dktc)

Answer 1

a) PTHH : 2Al + 3Cl2 => 2AlCl3

b) nAl= 1,35 / 27 = 0,05 mol

pthh => nAlCl3 = nAl = 0,05 mol

mAlCl3 = 0,05 x 133,5 = 6,675 g

pthh => nCl2 = 3/2 nAl = 3/2. 0,05 = 0,075 mol

VCl2 (đktc) = 0,075 x 22,4 = 1,68 l

Answer 2

n_Al = \(\dfrac{1,35}{27}\)= 0,05 (mol)

2Al + 3Cl₂ → 2AlCl₃ (t^o)

0,05.....0,075........0,05

⇒ m_AlCl3 = 0,05.133,5 = 6,675 (g)

⇒ V_Cl2 = 0,075.22,4 = 1,68 (l)