Chp1_BalEqns

# Chp1_BalEqns - 4 has four. Thus, if the stoichiometric...

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Let us consider balancing the following reaction: 6 2 4 4 3 H B NaBF NaBH BF + + (1) The easiest place to start is recognizing that the sodium goes from one compound in the reactants to one in the products and that both molecules have a single sodium atom. Thus, their stoichiometric coefficients must be 1:1. Let us call this number “x” for the moment: 6 2 4 4 3 H B NaBF x NaBH x BF + + (2) Now let us consider the fluorine. The fluorine goes from one compound in the reactants to one in the products, but BF 3 has three fluorine atoms, NaBF
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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....
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## This note was uploaded on 03/29/2009 for the course CH 201 taught by Professor Warren during the Fall '07 term at N.C. State.

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