# How would you construct a balanced equation for the reaction between X and water to form the hydroxide, XOH, and a gas?

Aug 27, 2016

Here's how I would do it.

#### Explanation:

Since the metal forms the hydroxide $\text{XOH}$, it must be an alkali metal.

That means the gas it creates is hydrogen, ${\text{H}}_{2}$.

The unbalanced equation is

${\text{X" color(white)(l)+ "H"_2"O" → "XOH" + "H}}_{2}$

Step 1. Start by putting a $1$ in front of the most complicated formula. Let's pick $\text{H"_2"O}$.

$\text{X" color(white)(l)+ color(red)(1)"H"_2"O" → "XOH" + "H"_2"O}$.

Step 2. Balance $\text{O}$.

Put a 1 in front of $\text{XOH}$.

${\text{X"color(white)(l) + color(red)(1)"H"_2"O" → color(blue)(1)"XOH" + "H}}_{2}$

Step 3. Balance $\text{H}$.

There is no way to balance $\text{H}$ without using fractions.

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

${\text{X"color(white)(l) + color(red)(2)"H"_2"O" → color(blue)(2)"XOH" + "H}}_{2}$

Now, we have fixed 4 $\text{H}$ atoms on the left and 2 on the right.

We can get 4 $\text{H}$ atoms on the right by putting a 1 in front of ${\text{H}}_{2}$.

${\text{X"color(white)(l) + color(red)(2)"H"_2"O" → color(blue)(2)"XOH" + color(orange)(1)"H}}_{2}$

Step 4. Finally, balance $\text{X}$.

Put a $2$ in front of $\text{X}$.

$\textcolor{g r e e n}{2} {\text{X" + color(red)(2)"H"_2"O" → color(blue)(2)"XOH" + color(orange)(1)"H}}_{2}$

The equation should now be balanced.

Step 5. Check that atoms balance.

$\text{Atom"color(white)(m)"LHS"color(white)(m)"RHS}$
stackrel(————————)(color(white)(m)"X"color(white)(mmm)2color(white)(mmll)2)
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m} 4 \textcolor{w h i t e}{m m l l} 4$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m} 2 \textcolor{w h i t e}{m m l l} 2$

All atoms balance.

The balanced equation is

$2 {\text{X" + 2"H"_2"O" → 2"XOH" + "H}}_{2}$