You follow a systematic procedure to balance the equation.

The other product is calcium hydroxide.

Start by converting the words to a chemical equation.:

#"CaC"_2 + "H"_2"O" → "C"_2"H"_2 + "Ca(OH)"_2#

A method that often works is first to balance everything other than #"O"# and #"H"#, then balance #"O"#, and finally balance #"H"#.

Another useful procedure is to start with what looks like the most complicated formula.

The most complicated formula looks like #"Ca(OH)"_2#. We put a 1 in front of it to remind ourselves that the number is now fixed.

#"CaC"_2 + "H"_2"O" → "C"_2"H"_2 + color(red)(1)"Ca(OH)"_2#

**Balance #"Ca"#:**

We have #"1 Ca"# on the right, so we need #"1 Ca"# on the left. We put a 1 in front of the #"CaC"_2#.

#color(orange)(1)"CaC"_2 + "H"_2"O" → "C"_2"H"_2 + color(red)(1)"Ca(OH)"_2#

**Balance #"C"#:**

We have fixed #"2 C"# on the left. We need #"2 C"# on the right. Put a 1 in front of #"C"_2"H"_2#.

#color(orange)(1)"CaC"_2 + "H"_2"O" → color(blue)(1)"C"_2"H"_2 + color(red)(1)"Ca(OH)"_2#

**Balance #"O"#:**

We have fixed #"2 O"# on the right. We need #"2 O"# on the left. Put a 2 in front of #"H"_2"O"#.

#color(orange)(1)"CaC"_2 + color(purple)(2)"H"_2"O" → color(blue)(1)"C"_2"H"_2 + color(red)(1)"Ca(OH)"_2#

Every formula now has a coefficient. We should have a balanced equation.

Let's check.

#bb("Atom" color(white)(m)"lhs"color(white)(m)"rhs")#

#color(white)(m)"Ca"color(white)(mml)1color(white)(mml)1#

#color(white)(m)"C"color(white)(mmm)2color(white)(mml)2#

#color(white)(m)"H"color(white)(mmm)4color(white)(mml)4#

#color(white)(m)"O"color(white)(mmm)2color(white)(mml)2#

All atoms balance. The balanced equation is

#"CaC"_2 + "2H"_2"O" → "C"_2"H"_2 + "Ca(OH)"_2#