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1 Chapter 12 Solutions 12.6 Solutions in Chemical Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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Presentation on theme: "1 Chapter 12 Solutions 12.6 Solutions in Chemical Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings."— Presentation transcript:

1 1 Chapter 12 Solutions 12.6 Solutions in Chemical Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2 2 Molarity in Chemical Reactions In a chemical reaction, The volume and molarity of a solution are used to determine the moles of a reactant or product. volume (L) x molarity ( mol ) = moles 1 L If molarity (mol/L) and moles are given, the volume (L) can be determined mol x 1 L = volume (L) mol

3 3 Using Molarity of Reactants How many mL of 3.00 M HCl are needed to react 4.85 g CaCO 3 ? 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) STEP 1 Given 3.00 M HCl; 4.85 g CaCO 3 Need volume in mL STEP 2 Plan g CaCO 3 mol CaCO 3 mol HCl mL HCl

4 4 Using Molarity of Reactants (cont.) 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) STEP 3 Equalitites 1 mol CaCO 3 = g; 1 mol CaCO 3 = 2 mol HCl 1000 mL HCl = 3.00 mol HCl STEP 4 Set Up 4.85 g CaCO 3 x 1 mol CaCO 3 x 2 mol HCl x 1000 mL HCl g CaCO 3 1 mol CaCO mol HCl = 32.3 mL HCl required

5 5 Learning Check How many mL of a M Na 2 S solution are needed to react 18.5 mL of M NiCl 2 solution? NiCl 2 (aq) + Na 2 S(aq) NiS(s) + 2NaClaq) 1) 4.16 mL 2) 6.24 mL 3) 27.8 mL

6 6 Solution 3) 27.8 mL L x mol NiCl 2 x 1 mol Na 2 S x 1000 mL 1 L 1 mol NiCl mol = 27.8 mL Na 2 S solution

7 7 Learning Check What is the molarity if 15.0 mL of AgNO 3 solution reacts with 22.8 mL of M MgCl 2 ? MgCl 2 (aq) + 2AgNO 3 (aq) 2AgCl(s) + Mg(NO 3 ) 2 (aq) 1) M 2) M 3) M

8 8 Solution 3) M AgNO L x mol MgCl 2 x 2 mol AgNO 3 x 1 = 1 L 1 mol MgCl L = mol/L = M AgNO 3

9 9 Learning Check How many liters of H 2 gas at STP are produced when 12.5 g Zn react with 20.0 mL of 1.50 M HCl? Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) 1) 4.28 L H 2 2) L H 2 3) L H 2

10 10 Solution 2) L H 2 Limiting Reactant 12.5 g Zn x 1.00 mol Zn x 1 mol H 2 x 22.4 L = 4.28 L H g Zn 1 mol Zn 1 mol L x 1.50 mol HCl x 1 mol H 2 x 22.4 L 1 L 2 mol HCl1 mol = L H 2 gas

11 11 Summary of Chemical Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings


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