# Question #7be8c

Apr 2, 2016

$P b + 2 H C l \to P {b}^{+ 2} C {l}_{2} + {H}_{2}$
Or
$P b + 4 H C l \to P {b}^{+ 4} C {l}_{4} + 2 {H}_{2}$

#### Explanation:

The answer depends on the lead's oxidation number. Most likely case would be the first, because it is easier to lose two electrons rather than four.

Apr 27, 2016

See below..

#### Explanation:

Lead + Hydrochloric Acid $\implies$ Lead Chloride + Hydrogen

Pb + HCl $\implies$ $P b C {l}_{2}$ + $H$
Note that this one in unbalanced.

To chemically balance it, you need to determine how many atoms are on the left and right hand side of the equals sign.

Let's take a closer look on Pb. There is one Pb on both left and right hand side of equals sign. So you don't have to bother about Pb - at this stage.

Now look at H. Both left and right hand side has one H atoms. So leave the H for now.

Look at Cl. On the left side, there is one, whereas on the right side, there is two. You can see the difference now. If the left has one Cl, you need one more Cl. So, now you have two Cl on each side. Add the 2 in front of the compound. Like that:
[Pb + 2HCl $\implies$ $P b C {l}_{2}$ + $H$ ]

Now check the equation again to see if it's completely balanced. Look carefully!

Pb has one atom on both sides - good.
Two Cl on both sides - good.
Two H on left side but One H on right side - fix it.

Since you have only one H on right of equals sign, you need to add another one - and it becomes 2. So add the 2 in front of the H: like that:
[Pb + 2HCl $\implies$ $P b C {l}_{2}$ + $2 H$]

Now check again just to be sure! Check it every time if you made a change.
Pb - one atom on both sides = good
H - two atoms on both sides = good
Cl - two atoms on both sides = good

[Pb + 2HCl $\implies$ $P b C {l}_{2}$ + $2 H$]