This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 4 has four. Thus, if the stoichiometric coefficient for BF 3 is y, we have the condition: 3y = 4x (to conserve the number of fluorine atoms) which implies: y = 4x/3 If we choose x = 3, then y = 4 and we have: 6 2 4 4 3 ? 3 3 4 H B NaBF NaBH BF + + (3) Now looking at the hydrogen, we find twelve hydrogen atoms on the LHS, thus we need a 2 for the stoichiometric coefficient of B 2 H 6 . 6 2 4 4 3 2 3 3 4 H B NaBF NaBH BF + + (4) Lastly, we confirm that the numbers of boron atoms are balanced; and they are with seven on each side....
View
Full
Document
This note was uploaded on 03/30/2008 for the course CH 201 taught by Professor Warren during the Fall '07 term at N.C. State.
 Fall '07
 Warren
 Chemistry, Reaction

Click to edit the document details