# Why can't I balance this? "CaOH" + "H"_2"SO"_4 -> "CaSO"_4 + "H"_2"O"

Jun 22, 2016

Well, your reaction is not real, so that would be why! You had the products correct, but check your reactants.

My result is:

$\textcolor{red}{2} \text{Ca"("OH")_2 (s) + color(red)(2)"H"_2"SO"_4 (aq) -> color(red)(2)"CaSO"_4 (s) + color(red)(4)"H"_2"O} \left(l\right)$

which simplifies to:

$\setminus m a t h b f \left(\text{Ca"("OH")_2 (s) + "H"_2"SO"_4 (aq) -> "CaSO"_4 (s) + color(red)(2)"H"_2"O} \left(l\right)\right)$

CHECKING OUR REACTANTS/PRODUCTS

You have ${\text{Ca}}^{2 +}$ ion (second column on the periodic table, not the first), but as you have written it, it is bonded with only one ${\text{OH}}^{-}$ (hydroxide) polyatomic ion.

So, its charge is not balanced, and you have defined an unstable or fake compound.

To fix that, simply use the correct compound that balances the charges:

"Ca"^(2+)(aq) + color(red)(\mathbf(2))"OH"^(-)(aq) -> color(black)("Ca"("OH")_color(red)(\mathbf(2))(s))

Also, you can see from the above that you have to use parentheses to specify that two polyatomic ions are being used here, not just one of those atoms in the ion itself.

As for sulfuric acid (${\text{H"_2"SO}}_{4}$), since ${\text{SO}}_{4}^{2 -}$ has a $- 2$ charge, and each $\text{H}$ ion has a $+ 1$ charge, the charge is balanced and this is a correct reactant that contains a polyatomic ion. Similarly, ${\text{CaSO}}_{4}$ is correct, and we know how water is written.

So, your correct starting reaction is:

$\textcolor{h i g h l i g h t}{\text{Ca"("OH")_2 (s) + "H"_2"SO"_4 (aq) -> "CaSO"_4 (s) + "H"_2"O} \left(l\right)}$

Now we can balance this properly.

BALANCING THE REACTION

I'm choosing to balance the calciums first, since there is only one species on each side of the reaction that contains calcium, making it easy to keep track of.

$\implies \textcolor{red}{2} \text{Ca"("OH")_2 (s) + "H"_2"SO"_4 (aq) -> color(red)(2)"CaSO"_4 (s) + "H"_2"O} \left(l\right)$

Now, the sulfates (${\text{SO}}_{4}^{2 -}$) are off-balance, so let's balance those by multiplying the number of ${\text{H"_2"SO}}_{4}$ (sulfuric acid) molecules.

$\implies \textcolor{red}{2} \text{Ca"("OH")_2 (s) + color(red)(2)"H"_2"SO"_4 (aq) -> color(red)(2)"CaSO"_4 (s) + "H"_2"O} \left(l\right)$

Our current number of hydrogens on each side is:

$2 \times \left(2 \times \text{H") + 2xx(2xx"H}\right)$ vs. $2 \times \text{H}$

So, we should quadruple the number of water molecules to get:

$\implies \textcolor{red}{2} \text{Ca"("OH")_2 (s) + color(red)(2)"H"_2"SO"_4 (aq) -> color(red)(2)"CaSO"_4 (s) + color(red)(4)"H"_2"O} \left(l\right)$

CHECKING THE ATOM COUNT + SIMPLIFICATIONS

Checking the final atom count, treating polyatomic ions like a collection of atoms. List to consider:

• $\text{Ca}$
• $\text{H}$
• $\text{O}$
• $\text{S}$

We have:

$2 \times \text{Ca}$ vs. $2 \times \text{Ca}$ color(green)(sqrt"")

$2 \times \left(2 \times \text{H") + 2xx(2xx"H}\right)$ vs. $4 \times \left(2 \times \text{H}\right)$ color(green)(sqrt"")

$2 \times \left(2 \times \text{O") + 2xx(4xx"O}\right)$ vs. 2xx(4xx"O") + 4xx"O" color(green)(sqrt"")

$2 \times \text{S}$ vs. $2 \times \text{S}$ color(green)(sqrt"")

So this is correct. Just one thing left to do. Since $2$ divides $4$, this simplifies to:

$\setminus m a t h b f \left(\textcolor{b l u e}{\text{Ca"("OH")_2 (s) + "H"_2"SO"_4 (aq) -> "CaSO"_4 (s) + color(red)(2)"H"_2"O} \left(l\right)}\right)$