Since the metal forms the hydroxide #"XOH"#, it must be an alkali metal.

That means the gas it creates is hydrogen, #"H"_2#.

The unbalanced equation is

#"X" color(white)(l)+ "H"_2"O" → "XOH" + "H"_2#

**Step 1.** Start by putting a #1# in front of the most complicated formula. Let's pick #"H"_2"O"#.

#"X" color(white)(l)+ color(red)(1)"H"_2"O" → "XOH" + "H"_2"O"#.

**Step 2.** Balance #"O"#.

Put a 1 in front of #"XOH"#.

#"X"color(white)(l) + color(red)(1)"H"_2"O" → color(blue)(1)"XOH" + "H"_2#

**Step 3.** Balance #"H"#.

There is no way to balance #"H"# without using fractions.

We start over, multiplying all our old coefficients by #2#.

#"X"color(white)(l) + color(red)(2)"H"_2"O" → color(blue)(2)"XOH" + "H"_2#

Now, we have fixed 4 #"H"# atoms on the left and 2 on the right.

We can get 4 #"H"# atoms on the right by putting a 1 in front of #"H"_2#.

#"X"color(white)(l) + color(red)(2)"H"_2"O" → color(blue)(2)"XOH" + color(orange)(1)"H"_2#

**Step 4.** Finally, balance #"X"#.

Put a #2# in front of #"X"#.

#color(green)(2)"X" + color(red)(2)"H"_2"O" → color(blue)(2)"XOH" + color(orange)(1)"H"_2#

The equation should now be balanced.

**Step 5.** Check that atoms balance.

#"Atom"color(white)(m)"LHS"color(white)(m)"RHS"#

#stackrel(————————)(color(white)(m)"X"color(white)(mmm)2color(white)(mmll)2)#

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

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

All atoms balance.

The balanced equation is

#2"X" + 2"H"_2"O" → 2"XOH" + "H"_2#