CHEM
Ex_1A_KEY.pdf

# Suppose 1050 g of solid caco 3 is added to 725 ml of

• Test Prep
• 8

This preview shows pages 7–8. Sign up to view the full content.

compound, so it dissociates to give calcium ions and chloride ions. Suppose 10.50 g of solid CaCO 3 is added to 725 mL of 0.225 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: 10.50 g CaCO 3 = 0.105 mol (MM = 100 g/mol) (0.725 L)(0.225 mol/L) = 0.163 mol HCl From those conclude that HCl is limiting. (various ways to do this) Then calculate from the HCl: ࠵?. ࠵?࠵?࠵? ࠵?࠵?࠵? ࠵?࠵?࠵? × ࠵? ࠵?࠵?࠵? ࠵?࠵?࠵?࠵? ࠵? ࠵? ࠵?࠵?࠵? ࠵?࠵?࠵? × ࠵? ࠵?࠵?࠵? ࠵?࠵? ࠵?Y ࠵? ࠵?࠵?࠵? ࠵?࠵?࠵?࠵? ࠵? = ࠵?. ࠵?࠵?࠵?࠵? ࠵?࠵?࠵? ࠵?࠵? ࠵?Y And that number of moles is in the 725 mL volume, so molarity is: 0.0815 mol Ca 2+ 0.725 L = ࠵?. ࠵?࠵?࠵? ࠵? ࠵?࠵? ࠵?Y If you chose the CaCl 2 as the LR, you would get 0.145 M Ca 2+

This preview has intentionally blurred sections. Sign up to view the full version.

NAME:______________________________________ © 2018 L.S. Brown A9 (10 pts) 19. KCl and KBr are both ionic solids. A mixture of KCl and KBr has a mass of 3.595 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 3.129 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 = 3.595 g – 3.129 g = 0.466 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: 0.466 g × 1 mol Br 44.45 g =0.0105 mol Br That’s also the moles of KBr initially present, so we can convert it to mass: 0.0105 mol KBr × 119.0 g KBr 1 mol KBr = 1.248 g KBr 1.248 g KBr 3.595 g total = 0.347 so the sample was 34.7% KBr An alternative approach would be to find the final mass if the original sample had been 100% KBr. 3.595 g KBr × 1 mol KBr 119.0 g × 1 mol KCl 1 mol KBr × 74.55 g KCl 1 mol KCl = 2.252 g So if it was 100% KBr, then we would have D m = 3.595 g – 2.252 g = 1.343 g Divide the actual D m of 0.466 g by that to get the fraction of KBr: 0.466 g 1.343 g = 0.347, which matches our previous result.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern