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tutorial 2 solutions - CHEM 1E03 TUTORIAL#2 Solutions 1...

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CHEM 1E03 TUTORIAL #2 Solutions _____________________________________________________________________ Page 1 1. Complete and balance the following reactions: write N.R. if no reaction occurs. For all reactions that yield products, classify the reactions as precipitation (PPT), oxidation-reduction (RED), acid-base (AB), or other. As well, for all of the redox reactions, select the oxidizing agent. a) Mg(s) + Br 2 (l) Mg(s) + Br 2 (l) MgBr 2 (s) This is a redox reaction (RED). Br 2 is the oxidizing agent. The product is an ionic solid. b) NaOH(aq) + H 2 SO 4 (aq) (complete reaction) 2 NaOH(aq) + H 2 SO 4 (aq) 2 Na + (aq) + SO 4 2 (aq) + 2 H 2 O(l) This is an acid-base reaction (AB). If all the water were driven off, Na 2 SO 4 (s) would result. c) H 2 S(aq) + Fe(NO 3 ) 2 (aq) H 2 S(aq) + Fe(NO 3 ) 2 (aq) FeS(s) + 2 H + (aq) + 2 NO 3 (aq) This is a precipitation reaction (PPT). The net ionic equation is: Fe 2+ (aq) + S 2 (aq) FeS(s) d) KCl(s) + H 2 O(l) KCl(s) + H 2 O(l) K + (aq) + Cl (aq) This equation represents the dissolving of a soluble salt. This reaction does not fit any of the three classes mentioned at the beginning of the question, but do not consider this reaction to be "NR" – it represents a dissolution process. e) P 4 (s) + O 2 (g)(excess) P 4 (s) + 5 O 2 (g) P 4 O 10 (s) This is an oxidation-reduction reaction (RED). O 2 is the oxidizing agent. Note that the O.N. of phosphorus changes from 0 to +5 here. Phosphorus is the reducing agent. With limited oxygen, the product is P 4 O 6 , in which phosphorus undergoes an O. N. change from zero to +3. f) Na(s) + H 2 O(l) 2 Na(s) + 2 H 2 O(l) 2 NaOH(aq) + H 2 (g) This is an oxidation-reduction reaction (RED). H + (from water) is the oxidizing agent; it undergoes an O.N. change from +1 to zero. Note that NaOH(aq) means Na + (aq) + OH (aq), since all sodium compounds are soluble in water. NaOH is a strong electrolyte.
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