Because, HCl (Hydrochloric acid) Is a VERY strong acid. It is soluble in very few things. Most things are soluble IN HCl. Not the other way around. The more acidic it its, the less soluble it is.
CH4 + Cl2 → CH3Cl + HCl - monohlor metan CH3Cl + Cl2 → CHCl2 + HCl - dihlor metan CHCl2 + Cl2 → CHCl3 + HCl - trihlor metan CHCl3 + Cl2 → CCl4 + HCl - tetrahlor metan
P4+Cl2 P4+Cl2
Cl2 + H2O = HOCl + HCl
For each mole of hydrogen gas (H2) reacting with chlorine gas (Cl2), you will get 2 moles of HCl. H2 + Cl2 = 2 HCl
The reaction is Cl2 + H2O --> HCl + HClO
CH4 + Cl2 → CH3Cl + HCl CH3Cl + Cl2 → CH2Cl2 + HCl CH2Cl2 + Cl2 → CHCl3 + HCl
You start with methane (CH4 ) and chlorine ( Cl2) and react them thus:- CH4 + Cl2 → CH3Cl + HCl CH3Cl + Cl2 → CH2Cl2 + HCl CH2Cl2 + Cl2 → CHCl3 + HCl
CH4 + Cl2 → CH3Cl + HCl - monohlor metan CH3Cl + Cl2 → CHCl2 + HCl - dihlor metan CHCl2 + Cl2 → CHCl3 + HCl - trihlor metan CHCl3 + Cl2 → CCl4 + HCl - tetrahlor metan
hcl
P4+Cl2 P4+Cl2
Cl2 + H2O = HOCl + HCl
H2 (g) + Cl2 (g) --> 2 HCl (g) 25.00 g HCl x 1 mol HCl x 1 mol Cl2 x 70.90 g Cl2 = 24.3 g Cl2 are needed. ................... 36.46 g HCl . 2 mol HCl .. 1 mol Cl2
3(Cl2) + 3(H2O) = 5(HCl) + 1(HClO3)
Sodium chloride is soluble only in the water solution of HCl.
Keq=[H2][Cl2]/[hcl]^2
Just replace one H and add a Cl (CH3Cl). In the next step replace another H by Cl. It's a chain reaction under sunlight: CH4+Cl2=CH3Cl+HCl CH3Cl+Cl2=CHCl2+HCL CH2Cl+Cl2=CHCl3+HCl CHCl3+Cl2=CCl4+HCl I'm sure now the structural formula will be apiece of cake.
NaClO3 + 6 HCl = 3 Cl2 + 3 H2O + NaCl