Question

is there a difference or is there something wrong with writing the equation like this: 2 Ca(s) + 2 H2O(l) = 2CaOH(aq) + H2(g)
instead of like this: Ca(s) + 2 H2O(l) = Ca(OH)2(aq) +H2(g)

Answer 1

Roku the formula on Ca(OH)2 is wrong in the first version. You have it right in the second version. Otherwise I don't see a difference.

Answer 2

More simply, the first reaction does not exist because Ca(OH) is not a known substance.

Answer 3

thanks @Vincent-Lyon.Fr

Answer 4

Welcome :)

Answer 5

so how can i know weather or not something exists like that or not... cause im having trouble like this in other equation i write.

Answer 6

Calcium is an element in the second group of the periodic table.
It will have only one stable ion which is Ca2+, as it will lose its last 2 electrons and have stable state as argon.

OH- has only one extra negative charge.

In order to have a neutral compound, you need 2 hydroxide ions for each calcium ion, hence the formula Ca(OH)2.

If you work with potassium instead, which is in group 1, then ion is K+ and potassium hydroxide is KOH

Answer 7

these are all ionic bonding right?

Answer 8

Yes they are.
Actually, OH- being a polyatomic anion has a covalent bond between O and H too.

What we wrote would hold for CaCl2 and KCl with anion Cl- which is monatomic.

Answer 9

so the only way to seperate these two reactions again would be by electolysis.
why is OH covalent?

Answer 10

What do you mean "separate"?

Electrolysis is not just separating Ca2+ and Cl- (as in 'bringing them apart'), it would actually change their chemical nature by oxidizing one and reducing the other.

Answer 11

i still dont quite understand oxidizing and reduction
i just know oxidizing is the add of oxygen and removal of hydrogen while reduction is the opposite of the oxidation.

Answer 12

Actually adding electrons is also reduction: \(Ca^{2+}+2e^-\rightarrow Ca \)

Removing electrons is oxidation: \( Cl^- \rightarrow Cl+e^-\)

Answer 13

electrolisys cause ions to form back into their element isnt that just about accurate?

Answer 14

It is not always the case, but it is mostly used to achieve what you say.

Answer 15

For instance, in aqueous solution, you cannot perform:
\( Na^+ + e^-\rightarrow Na\)
because Na+ is too difficult to reduce and water gets reduced instead.

Answer 16

my mind is still a bit slow in processing this but ok thanks @Vincent-Lyon.Fr