# Question On Balancing [closed]

I recently worked on a lab which involved the following reaction:

$$\ce{HCl(aq) + CuO(s) -> CuCl2(aq) + H2O(l)}$$

In my description of the reaction I wrote the following

"The equation shown above represents a multi–step reaction between copper (II) oxide and hydrochloric acid in aqueous solution. To predict the products, we must recall that a metal oxide turns into a metal hydroxide when placed in water. The initial copper ion had a +2 charge, so the hydroxides must have a net charge of –2. This is accomplished through two hydroxyl groups. Once copper (II) hydroxide has formed, it may react with hydrochloric acid. These reactants undergo a special type of double replacement reaction known as a neutralization reaction. In essence, the metals combine ($$\ce{CuCl2}$$) and water is formed"

From here, I was wondering how exactly to balance the equation. Specifically, do I have

1. $$\ce{CuO(s) + 2 HCl(aq) -> Cu(OH)2(aq) + 2 HCl(aq) -> CuCl_2(aq) + 2 H2O(l)}$$
2. $$\ce{CuO(s) + 2 HCl(aq) -> Cu(OH)2(aq) + 2 HCl(aq) -> CuCl2(aq) + H2O(l)}$$

It seems like I can "cancel out" a water molecule when going from middle reactants to final reactants. Is this true?

• CuO is very poorly soluble in water and there is no intermediate copper hydroxide formation in HCl solution. The proposed description looks like a Rube Goldberg machine to me. Do you have any experimental evidences to support suggested "mechanism"? – andselisk Jan 5 '20 at 20:51
• Your equations are not balanced (count the atoms!). First balance, then cancel out. Then you'll get a single result. Besides that, as @andselisk said, there's no reason to assume intermediate copper hydroxide. – aventurin Jan 7 '20 at 20:03