How do you balance this chemical equation: #"Mg"_3"B"_2 + "H"_2"O" - "Mg"("OH")_2 + "B"_2"H"_6#?

Question

How do you balance this chemical equation: #"Mg"_3"B"_2 + "H"_2"O" -> "Mg"("OH")_2 + "B"_2"H"_6#?

Naomi Aronson · Accepted Answer

What I received:

#"Mg"_3"B"_2 + color(red)(6)"H"_2"O" -> color(red)(3)"Mg"("OH")_2 + "B"_2"H"_6#

We begin with:

#"Mg"_3"B"_2 + "H"_2"O" -> "Mg"("OH")_2 + "B"_2"H"_6#

and determine which atoms are out of balance first.

In order to balance the magnesium, oxygen, or hydrogen, but not the boron, I would therefore add whole-number coefficients.

Since the boron is already balanced, I would ignore the #"B"_2"H"_6# and start with tripling the quantity of #"Mg"("OH")_2#.

#"Mg"_3"B"_2 + "H"_2"O" -> color(red)(3)"Mg"("OH")_2 + "B"_2"H"_6#

The magnesium is now in balance.

There are #2xx3 = color(red)(6)# oxygen atoms on the products side, so we multiply the water molecules by #\mathbf(6)# on the reactants side to get #color(red)(6)# oxygen atoms on the reactants side.

#color(blue)("Mg"_3"B"_2 + color(red)(6)"H"_2"O" -> color(red)(3)"Mg"("OH")_2 + "B"_2"H"_6)#

Presently, we possess:

This is balanced because the atoms of magnesium, oxygen, and hydrogen have all been balanced without affecting the atoms of boron, which were already balanced.

Noah Aldrich · Answer

undefined Balanced equation: $3 \text{Mg}_3\text{B}_2 + 6 \text{H}_2\text{O} \rightarrow 3 \text{Mg}(\text{OH})_2 + 2 \text{B}_2\text{H}_6$

How do you balance this chemical equation: #"Mg"_3"B"_2 + "H"_2"O" -&gt; "Mg"("OH")_2 + "B"_2"H"_6#?

How do you balance this chemical equation: #"Mg"_3"B"_2 + "H"_2"O" -> "Mg"("OH")_2 + "B"_2"H"_6#?