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

1 Answer
Truong-Son N.
Jun 26, 2016

I got:

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

We start with:

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

and first see which atoms are unbalanced.

  • #stackrel("Reactants Side")overbrace([3xx"Mg"])# vs. #stackrel("Products Side")overbrace([1xx"Mg"])# --- unbalanced
  • #stackrel("Reactants Side")overbrace([2xx"B"])# vs. #stackrel("Products Side")overbrace([2xx"B"])# --- balanced
  • #stackrel("Reactants Side")overbrace([2xx"H"])# vs. #stackrel("Products Side")overbrace([2xx"H" + 6xx"H"])# --- unbalanced
  • #stackrel("Reactants Side")overbrace([1xx"O"])# vs. #stackrel("Products Side")overbrace([2xx"O"])# --- unbalanced

So, I would add whole-number coefficients to balance either the magnesium, oxygen, or hydrogen, but not the boron.

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#

Now the magnesium are balanced.

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)#

Now we have:

  • #2xx6 = color(red)(12)# hydrogen atoms on the reactants side, and #3xx2 + 6 = color(red)(12)# hydrogen atoms on the products side, which is balanced.
  • #color(red)(6)# oxygen atoms on the reactants side, and #3xx2 = color(red)(6)# oxygen atoms on the products side, which is balanced.

Since we have balanced the magnesium, oxygen, and hydrogen atoms without touching the boron atoms (which were already balanced), this is balanced.

  • #stackrel("Reactants Side")overbrace([3xx"Mg"])# vs. #stackrel("Products Side")overbrace([3xx"Mg"])# --- balanced
  • #stackrel("Reactants Side")overbrace([2xx"B"])# vs. #stackrel("Products Side")overbrace([2xx"B"])# --- balanced
  • #stackrel("Reactants Side")overbrace([6xx2xx"H"])# vs. #stackrel("Products Side")overbrace([3xx2xx"H" + 6xx"H"])# --- balanced
  • #stackrel("Reactants Side")overbrace([6xx"O"])# vs. #stackrel("Products Side")overbrace([3xx2xx"O"])# --- balanced
Related questions
See all questions in Balancing Chemical Equations
Impact of this question
1508 views around the world
You can reuse this answer
Creative Commons License