Question

Cần bao nhiêu KMnO₄ và bao nhiêu mL dd HCL 1M để điều chế được 3,36 lít khí cl₂ (đktc) .

Answer 1

Answer 2

nCl2 = 0.15 mol

2KMnO4 + 16HCl --> 2KCl + 2MnCl2 + 5Cl2 + 8H2O

0.06________0.48___________________0.15

mKMnO4 = 0.06*158=9.48 g

Vdd HCl = 0.48 (l) = 480 ml

Answer 3

\(n_{Cl_2}=\frac{V}{22,4}=\frac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:16HCl+2KMnO_4\rightarrow5Cl_2+8H_2O+2KCl+2MnCl_2\\ m_{KMnO_4}=0,06.158=9,48\left(g\right)\\ V_{HCl}=\frac{n}{C_M}=\frac{0,48}{1}=0,48\left(l\right)\)

Answer 4

\(n_{Cl_2}=\frac{3,36}{22,4}=0,15mol\)

PTHH : 2KMnO₄ + 16HCl --> 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

(mol ) 0.06 0.48 0.15

\(m_{KMnO_4}=0,06.158=9,48g\)

=> Vdd HCl = 0,48 . 1 = 0,48 l