In the chemical equation MnO₂ + xHCl → MnCl₂ + yH₂O + zCl₂, what are the values of x, y, and z?

What you are solving here

You are finding the coefficients $x$, $y$, and $z$ that make the number of atoms of each element the same on both sides of the equation.

Balance atoms other than H and Cl first

Start with $\mathrm{MnO_2}$.

• Mn: there is 1 Mn on the left, so put 1 $\mathrm{MnCl_2}$ on the right (already 1 Mn).
• O: there are 2 O on the left, so you need $2\mathrm{H_2O}$ on the right to have 2 O.

Now the equation looks like:

$$ \mathrm{MnO_2 + xHCl \rightarrow MnCl_2 + 2H_2O + zCl_2} $$

Use hydrogen to determine $x$

On the right, $2\mathrm{H_2O}$ contains $4$ H atoms. So you need $4\mathrm{HCl}$ on the left to supply 4 H atoms, which gives:

$$x = 4$$

Balance chlorine to find $z$

With $4\mathrm{HCl}$, the left side has 4 Cl atoms. On the right:

• $\mathrm{MnCl_2}$ has 2 Cl atoms
• $\mathrm{Cl_2}$ has $2z$ Cl atoms

So:

$$ 2 + 2z = 4 \Rightarrow 2z = 2 \Rightarrow z = 1 $$

Also, from earlier, $y = 2$.

Final check

Balanced equation:

$$ \mathrm{MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2} $$

Counts: Mn $1=1$, O $2=2$, H $4=4$, Cl $4=2+2$.