Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 3 - Mass Relationships in Chemical Reactions - Section Problems - Page 106: 37


(a) Not balanced. The balanced equation is: $2Al + Fe_2O_3 -- \gt Al_2O_3 + 2Fe$ (b) It is already balanced. (c) Not balanced. $4Au + 8NaCN + O_2 + 2H_2O --\gt 4NaAu(CN)_2 + 4NaOH$

Work Step by Step

(a) $Al + Fe_2O_3 -- \gt Al_2O_3 + Fe$ The $Al$ compounds are not balanced: put a 2 on the one on the left side. $2Al + Fe_2O_3 -- \gt Al_2O_3 + Fe$ Now, the only non balanced element is $Fe$. We can adjust that by putting a 2 on the $Fe$ on the right side. $2Al + Fe_2O_3 -- \gt Al_2O_3 + 2Fe$ (b) The number of carbons is equal to 6 on both sides, that for hydrogen is 12 on both sides, and we have 18 oxygens on both sides. (c) Balance the $CN$ compounds. $Au + 2NaCN + O_2 + H_2O --\gt NaAu(CN)_2 + NaOH$ Balance the hydrogen number: $Au + 2NaCN + O_2 + H_2O --\gt NaAu(CN)_2 + 2NaOH$ We need to add 1 $Na$ to the left side, but the $CN$ number has to still the same, we can do that by doing this: $Au + 4NaCN + O_2 + H_2O --\gt 2NaAu(CN)_2 + 2NaOH$ Balance $Au:$ $2Au + 4NaCN + O_2 + H_2O --\gt 2NaAu(CN)_2 + 2NaOH$ Now, the only unbalanced element is oxygen; trying to balance it: $Au + 4NaCN + \frac{1}{2}O_2 + H_2O --\gt 2NaAu(CN)_2 + 2NaOH$ Multiply all coefficients by 2. $4Au + 8NaCN + O_2 + 2H_2O --\gt 4NaAu(CN)_2 + 4NaOH$
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