Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: charlie1983 on July 31, 2006, 11:28:20 PM
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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
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I STILL count 3 hydrogens on the left side and 4 hydrogens on the right side.
Count harder ;)
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
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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.