Question On Balancing [closed]

Question

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?

Answer 1

As andselisk points out, CuO is poorly soluble, but the hydroxide is listed in the CRC Handbook (blue, insoluble, no numbers), so it can be assumed to exist under some conditions. The Reference Book of Inorganic Chemistry, Latimer and Hildebrand, 3rd ed, 1951, p110, mentions Cu(OH)2 being formed in solution by addition of hydroxide, cold, to Cu++ as green or bluish-green (in hot solution, black CuO is formed). Cu(OH)2 is a very weak base; K = 5.6 e-20 (this must be Ksp).

Now, your experiment: you start with black CuO and wind up with green CuCl2. Because the blue/bluish-green Cu(OH)2 color will be so faint because of its low solubility, I'm going to suggest that under ordinary concentrations of HCl (~0.1 - 1 M), you will not be able to detect the very fast appearance and disappearance of Cu(OH)2 in solution. However, with a large excess of CuO and very dilute HCl, you might, under some conditions, be able to get some Cu(OH)2 into solution if you did it delicately. But I wouldn't think this would be easy. If you wanted to see Cu(OH)2, the way to go would be to add cold NaOH solution to a cold copper salt solution.

Now, for your equations: #1 is not balanced: you have two total hydrogens on the left and four on the right. #2 is balanced right and left, but the middle has too many hydrogens.

I think you overthought the reaction. It might be two-step, but that intermediate step is very difficult to investigate and can be swept under the carpet. But its better to overthink and walk back than to underthink and miss the boat.