HCHO2 is a weird way of writing formic acid, which is usually written CHOOH. As you might be able to guess from it's name, it's an acid. KOH is a base. Like most acids and bases, they react to make water, and a salt. The salt in this case would be potassium formate.
'HCHO2' m which should be written as 'HCOOH'. is methanoic acid.
Being an acid it will react with an alkali.
Hence
HCOOH + KOH = HCOO^-K^+ + H2O
HCOO^-K^+ is potassium methanoate.
HCHO2 is methanoic acid and is normally written as HCOOH. When reacted with potassium hydroxide (KOH) , it produces potassium methanoate. and water/ HCOOH +KOH = HCOO^-K+ + H2O
C2h4
It is a strong acid
KOH act as a nucleufeel and react with benzil OH give electrones to benzil and berak the double bond.
They do not react, the Calcium cannot displace the Potassium as it is less reactive.
HCHO2 is methanoic acid and is normally written as HCOOH. When reacted with potassium hydroxide (KOH) , it produces potassium methanoate. and water/ HCOOH +KOH = HCOO^-K+ + H2O
C2h4
It is a strong acid
KOH act as a nucleufeel and react with benzil OH give electrones to benzil and berak the double bond.
They do not react, the Calcium cannot displace the Potassium as it is less reactive.
Nothing, they don't react with each other.
I assume KOH is limiting. Balanced equation. KOH + HCl -> KCl + H2O 0.400 moles KOH (1 mole H2O/1 mole KOH)(18.016 grams/1 mole H2O) = 7.21 grams water produced =====================
When alc. KOH react with alkyle halide it for Alkene, KX (X Is stande for halide) and water. And this reaction also called Dehydrohalogenation...
From the balanced equation it can be seen that it takes 2 moles KOH to react with each 1 mole of MgCl2. So, the answer is 2.
No. KH (potassium hydride) is a base. It will react with water in this manner: KH + H2O --> KOH + H2
MgCl2 + 2KOH ==> Mg(OH)2 + 2KCl1 mole MgCl2 reacts with 2 moles KOH 2 moles KOH x 56.1 g/mole = 112.2 g KOH = 100 g KOH (to 1 significant figure based on 1 mole)
There will be no reaction. These two compounds do not react with each other at all.