hmwk4 - Chemistry 209 Chapter 4: Homework #4 answers 4 (a)...

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Chemistry 209 Homework #4 answers Chapter 4: 4 (a) Strategy: what’s the easiest thing to keep track of? The most complicated item; that is the starting material SO 2 Cl 2 , as any changes in the amounts of that stuF require that I rebalance 3 of the 4 product molecules. HI I will worry about last, as it looks like it is the source of H atoms that appear in everything on the right except the I 2 ;atomso fI , because they are really just spectators to the rest of the reaction I’ll worry about last when we have to count up how many spectators there are at this reaction. So we’ll start assuming that there is a single unit of the SO 2 Cl 2 , as see how much trouble that gets us in; as a ±rst guess, I’ll write down SO 2 Cl 2 + ° HI H 2 S+2H 2 O+2HCl+ ° I 2 where ° represents a coefficient to be determined later. Like, now; counting up the H atoms on the right side, I’ve got 2 in the H 2 S molecule, 4 in the two H 2 O molecules, and 2 more in the 2 HCl molecules, for a total of 8; that means that I need 8 units of HI (my only source for H) and that leaves me with 8 units of I, or 4 I 2 molecules, as passengers along for the ride. Ending with SO 2 Cl 2 +8HI H 2 S+2H 2 O+2HCl+4I 2 (b) Strategy: Count what we have no other source for ±rst; that would be the ²eTiO 3 ;one unit of which must yield one unit of each of the two product species. So start with ²eTiO 3 + ° H 2 SO 4 + ° H 2 O ²eSO 4 7H 2 O+TiOSO 4 Next, we need enough of the SO 4 units to go around; that looks like two on the right, so we need two on the left, giving ²eTiO 3 +2H 2 SO 4 + ° H 2 O ²eSO 4 7H 2 O+TiOSO 4 Now it’s time to ask whether we have atom balances that the added water is meant to ±x up; the ²e, Ti and S atoms are all in balance on both sides of the equation; on the product side we have a total of 14 H atoms and 16 O atoms. So far the reactants contribute 3 + 2 × 4 = 11 O atoms, and 4 H atoms; so we need 10 more H atoms and 5 more O atoms on the reactant side–which conveniently enough is provided by exactly 5 water molecules, leading to ²eTiO 3 +2H 2 SO 4 +5H 2 O ²eSO 4 7H 2 O+TiOSO 4 (d) This one is a simple combustion problem; C, H and S burn in O to form CO 2 ,H 2 Oand SO 2 . Simpler if we rewrite the ±rst molecule as C 14 H 14 S 2 ;then C 14 H 14 S 2 + ° O 2 →° CO 2 + ° H 2 O+ ° SO 2 Sequentially, there is only one source of C, and one product containing C; so all 14 C units need to be found in 14 CO
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This homework help was uploaded on 01/31/2008 for the course CHEM 2090 taught by Professor Zax,d during the Fall '07 term at Cornell University (Engineering School).

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hmwk4 - Chemistry 209 Chapter 4: Homework #4 answers 4 (a)...

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