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 stuff 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 ; atoms of I, 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 first guess, I’ll write down SO 2 Cl 2 + HI H 2 S + 2 H 2 O + 2 HCl + 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 + 8 HI H 2 S + 2 H 2 O + 2 HCl + 4 I 2 (b) Strategy: Count what we have no other source for first; that would be the FeTiO 3 ; one unit of which must yield one unit of each of the two product species. So start with FeTiO 3 + H 2 SO 4 + H 2 O FeSO 4 7 H 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 FeTiO 3 + 2 H 2 SO 4 + H 2 O FeSO 4 7 H 2 O + TiOSO 4 Now it’s time to ask whether we have atom balances that the added water is meant to fix up; the Fe, 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 FeTiO 3 + 2 H 2 SO 4 + 5 H 2 O FeSO 4 7 H 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 O and SO 2 . Simpler if we rewrite the first 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
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