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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: charlie1983 on July 31, 2006, 11:28:20 PM

Title: Writing an Equation for a described Reaction
Post by: charlie1983 on July 31, 2006, 11:28:20 PM
Thanks in advance!

Directions:  Write an equation for the "described" reaction below.

Problem:   Bubbles form when a metallic barium is placed in water

What I did:   Ba (s) + HOH (aq) >>>Ba(OH)2 (aq) + H2

Balance the Equation:   Ba (s) + "2" HOH (aq) >>>Ba(OH)2 (aq) + H2

Question:  Are the HYDROGENS really BALANCED?  I STILL count 3 hydrogens on the left side and 4 hydrogens on the right side. Please explain the balancing of the hydrogens shown with as much detail as you can.

Thank you very much

Title: Re: Writing an Equation for a described Reaction
Post by: Borek on August 01, 2006, 03:19:19 AM
I STILL count 3 hydrogens on the left side and 4 hydrogens on the right side.

Count harder ;)

Quote
Please explain the balancing of the hydrogens shown with as much detail as you can.

2 particles of water (H2O, that you for some reason decided to write as HOH) containing 2 atoms of hydrogen each. 2*2= ?

http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-reactions
Title: Re: Writing an Equation for a described Reaction
Post by: sdekivit on August 01, 2006, 06:28:31 AM
Thanks in advance!


Balance the Equation:   Ba (s) + "2" HOH (aq) >>>Ba(OH)2 (aq) + H2

this is the correct equation. A general way to balance equations is solving a set of equations:

a Ba + b H2O --> c Ba(OH)2 + d H2

Now:

Ba --> a = c
H --> 2b = 2c + 2d
O --> b = 2c

Now use substitution to get b = 2c

Now all variables are dependent on c thus we take c = 1 for the lowest coefficients. It automatically follows a = 1, b = 2, c = 1 and d = 1.