# Question #6cbbb

Feb 25, 2018

$C u O \left(s\right) + 2 H C l \left(a q\right) \to C u C {l}_{2} \left(a q\right) + {H}_{2} O \left(l\right)$

#### Explanation:

This is a neutralization reaction. In a neutralization reaction, the chemical equation is as follows:

$\text{acid + base" \ -> \ "salt + water}$

Here, we got $C u O$ as a base, since it can react with water to form $C u {\left(O H\right)}_{2}$, which is a basic solution. The acid here is $H C l$.

So, our reaction will be

$C u O \left(s\right) + H C l \left(a q\right) \to C u C {l}_{2} \left(a q\right) + {H}_{2} O \left(l\right)$

To balance it, I see $2$ chlorines on the right side, while only one on the left side, so I multiply $H C l$ by $2$. This gives us:

$C u O \left(s\right) + 2 H C l \left(a q\right) \to C u C {l}_{2} \left(a q\right) + {H}_{2} O \left(l\right)$

This is also a reaction that has been covered here:

https://socratic.org/questions/cuo-s-hcl-aq-equation-and-net-ionic-equation

Feb 25, 2018

$\text{CuO" + "2HCl" rarr "CuCl"_2+"H"_2"O}$

#### Explanation:

It is a double displacement reaction, where positive charged elements change their negative pairs.

What I mean is:
$\text{Cu"^(+2) "O"^(-2)+ "H"^(+1)"Cl"^(-1) rarr "CuCl"_2 + "OH}$

But we need to balance it so:
$\text{CuO" + 2"HCl" rarr "CuCl"_2 + "H"_2"O}$

This happens because the products have an extra Chlorine ($\text{Cl}$) so we multiply $\text{HCl}$ by $2$ and, therefore, add a Hydrogen ($\text{H}$) to the products, as well.