S brown 10 pts 18 calcium carbonate limestone caco 3

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© 2018 L.S. Brown (10 pts) 18. Calcium carbonate (limestone, CaCO 3 ) dissolves in hydrochloric acid, producing water and carbon dioxide as shown in the following equation. Calcium chloride is a soluble ionic compound, so it dissociates to give calcium ions and chloride ions. Suppose 8.50 g of solid CaCO 3 is added to 725 mL of 0.150 M HCl. Assuming that the volume of the final solution is still 725 mL, what concentration of Ca 2+ ions should be present? CaCO 3 (s) + 2 HCl(aq) ® H 2 O( ! ) + CO 2 (g) + CaCl 2 (aq) We are given the amounts of each reactant, so we need to start by finding the limiting reactant. Convert given info to moles: 8.50 g CaCO 3 = 0.085 mol (MM = 100 g/mol) (0.725 L)(0.15 mol/L) = 0.10875 mol HCl From those conclude that HCl is limiting. (various ways to do this) Then calculate from the HCl: And that number of moles is in the 725 mL volume, so molarity is: If you chose the CaCl 2 as the LR, you would get 0.117 M Ca 2+ 0.10875 mol HCl × 1 mol CaCl 2 2 mol HCl × 1 mol Ca 2 + 1 mol CaCl 2 = 0.0544 mol Ca 2 + 0.0544 mol Ca 2 + 0.725 L = 0.0750 M Ca 2 +
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NAME:______________________________________ © 2018 L.S. Brown B9 (10 pts) 19. KCl and KBr are both ionic solids. A mixture of KCl and KBr has a mass of 5.395 g. When this mixture is heated in the presence of excess Cl 2 , all of the KBr is converted to KCl. If the total mass of KCl present after this reaction is 4.085 g, what percentage (by mass) of the original mixture was KBr? (HINT: Be sure that you understand why the mass of the sample has decreased. It may help if you write an equation for the reaction that converted the KBr to KCl.) The mass decreases because the reaction replaces a heavier Br with a lighter Cl. There are a couple of ways we could work the problem, but no matter what we do we will need to use the change in mass, which is just D m = 5.395 g – 4.085 g = 1.310 g Maybe the easiest thing to do is to then say that if we had replaced 1 mole of Br with Cl, the change in mass would just be the difference in the molar masses: m Br – m Cl = 79.90 – 35.45 = 44.45 g/mol We can use those two numbers to find the number of moles of Br actually replaced: 1.310 g × 1 mol Br 44.45 g = 0.0295 mol Br That’s also the moles of KBr initially present, so we can convert it to mass: 0.0295 mol KBr × 119.0 g KBr 1 mol KBr = 3.507 g KBr 3.507 g KBr 5.395 g total =0.650 so the sample was 65.0% KBr An alternative approach would be to find the final mass of KCl if the original sample had been 100% KBr. 5.395 g KBr × 1 mol KBr 119.0 g × 1 mol KCl 1 mol KBr × 74.55 g KCl 1 mol KCl = 3.380 g So if it was 100% KBr, then we would have D m = 5.395 g – 3.380 g = 2.015 g Divide the actual D m of 1.310 g by that to get the fraction of KBr: 1.310 g 2.015 g =0.650, which matches our previous result.
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