How can I balance the following?
___Cu(OH)2 + ____K2SO4 = ___KOH + ____CuSO4

Question

How can I balance the following?
___Cu(OH)2 + ____K2SO4 = ___KOH + ____CuSO4

mathstudent55 · Answer

$ \_\_\_ Cu(OH)_2 + \_\_\_K_2SO_4 = \_\_\_KOH + \_\_\_CuSO_4$

Look at Cu.
There is 1 atom on each side, so it's balanced.

Now look at $O$. There are 2 atoms in $(OH)_2$ plus 4 atoms in $K_2SO_4$ on the left side.
That is 6 atoms on the left side. We need 6 atoms on the right side.

The right side has 1 $O$ atom in $KOH$ plus 4 more $O$ atoms in $CuSO_4$ for a total of 5.
If we have 2 molecules of $KOH$, we'll have the extra atom of $O$ we need on the right side.

$ \_\_\_ Cu(OH)_2 + \_\_\_K_2SO_4 = \color{red}{2}KOH + \_\_\_CuSO_4$

mathstudent55 · Answer

Now we start checking again.
Cu is still balanced with one atom on each side.
We have 2 O + 4 O on the left side and 2 O + 4 O on the right side.
We have 2 H on the left side and 2 H on the right side.
There are 2 K on the left and 2 K on the right.
Finally there is 1 S on the left and 1 S on the right.
The equation is balanced.

anonymous · Answer

I see, so would it be 1 Cu(OH)2 + 1K2SO3 = 2KOH + 1CuSO4?

mathstudent55 · Answer

Don't write 1's.
Just like in algebra we write simply x, not 1x, if the multiplier is 1, just leave it out.

anonymous · Answer

Oh okay, thank you very much, your explanation was incredibly helpful :)

anonymous · Answer

Wait, would this be a double replacement type of equation ?

mathstudent55 · Answer

I think so.

anonymous · Answer

I'm truly not sure. I am very bad at determining which reactions are which when I don't see them.

anonymous · Answer

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