1. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 12 Solutions 12.6 Solutions in Chemical Reactions When a BaCl 2 solution is added to a Na 2 SO 4 solution, BaSO 4, a white solid, forms.

2. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 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. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 Calculations Involving Solutions in Chemical Reactions

4. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 4 Example of Using Molarity in a Chemical Equation How many milliliters of a 3.00 M HCl solution are needed to react with 4.85 g of CaCO 3 ? 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) STEP 1 State the given and needed quantities. Given 3.00 M HCl solution; 4.85 g of CaCO 3 Need volume in milliliters STEP 2 Write a plan to calculate needed quantity or concentration. grams of CaCO 3 moles of CaCO 3 moles of HCl milliliters of HCl

5. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 Example of Using Molarity in a Chemical Equation (continued) STEP 3 Write equalities and conversion factors including mole-mole and concentration factors. 1 mol of CaCO 3 = 100.09 g of CaCO 3 1 mol CaCO 3 and 100.09 g CaCO 3 100.09 g CaCO 3 1 mol CaCO 3 1 mol of CaCO 3 = 2 mol of HCl 1 mol CaCO 3 and 2 mol HCl 2 mol HCl 1 mol CaCO 3 1000 mL of HCl solution = 3.00 mol of HCl 1000 mL HCl solution and 3.00 mol HCl 3.00 mol HCl 1000 mL HCl solution

6. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 Example of Using Molarity in a Chemical Equation (continued) STEP 4 Set up problem to calculate needed quantity or concentration. 4.85 g CaCO 3 x 1 mol CaCO 3 x 2 mol HCl 100.09 g CaCO 3 1 mol CaCO 3 x 1000 mL HCl = 32.3 mL of HCl solution 3.00 mol HCl

7. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 Learning Check How many milliliters of a 0.150 M Na 2 S solution are needed to react with 18.5 mL of a 0.225 M NiCl 2 solution? NiCl 2 (aq) + Na 2 S(aq) NiS(s) + 2NaCl(aq) A. 4.16 mL B. 6.24 mL C. 27.8 mL

8. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 Solution STEP 1 State the given and needed quantities. Given 0.0185 L of a 0.225 M NiCl 2 solution; 0.150 M Na 2 S solution Need milliliters of Na 2 S solution STEP 2 Write a plan to calculate needed quantity or concentration. liters of NiCl 2 solution moles of NiCl 2 solution moles of Na 2 S solution milliliters of Na 2 S solution

9. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 Solution (continued) STEP 3 Write equalities and conversion factors including mole-mole and concentration factors. 0.225 mol of NiCl 2 = 1 L of NiCl 2 solution 0.225 mol NiCl 2 and 1 L NiCl 2 1 L NiCl 2 0.225 NiCl 2 1 mol of NiCl 2 = 1 mol of Na 2 S 1 mol NiCl 2 and 1 mol Na 2 S 1 mol Na 2 S 1 mol NiCl 2 1000 mL of Na 2 S solution = 0.150 mol of Na 2 S 1000 mL HCl and 0.150 mol HCl 0.150 mol HCl 1000 mL HCl

10. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 Solution (continued) STEP 4 Set up problem to calculate needed quantity or concentration. 0.0185 L x 0.225 mol NiCl 2 x 1 mol Na 2 S x 1000 mL 1 L 1 mol NiCl 2 0.150 mol = 27.8 mL of Na 2 S solution (C)

11. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 11 Learning Check How many liters of H 2 gas at STP are produced when 6.25 g of Zn react with 20.0 mL of a 1.50 M HCl solution? Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) A. 4.28 L of H 2 B. 0.336 L of H 2 C. 0.168 L of H 2

12. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution STEP 1 State the given and needed quantities. Given 6.25 g of zinc; 0.0200 L of a 1.50 M HCl solution Need L of H 2 gas at STP STEP 2 Write a plan to calculate needed quantity or concentration. Limiting reactant: lowest number of moles of H 2 1) grams of zinc moles of zinc moles of H 2 2) liters of HCl solution moles of HCl solution moles of H 2 12

13. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 13 Solution (continued) STEP 3 Write equalities and conversion factors including mole-mole and concentration factors. 1 mol of Zn = 65.41 g of Zn 1 mol Zn and 65.41 g Zn 65.41 g Zn 1 mol Zn 1 mol of Zn = 1 mol of H 2 1 mol Zn and 1 mol H 2 1 mol H 2 1 mol Zn 2 mol of HCl = 1 mol of H 2 2 mol HCl and 1 mol H 2 1 mol H 2 2 mol HCl

14. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 14 Solution (continued) STEP 3 (continued) 1 mol of H 2 = 22.4 L of H 2 22.4 L H 2 and 1 mol H 2 1 mol H 2 22.L H 2 1.50 mol of HCl = 1 L of HCl solution 1.50 mol HCl and 1 L HCl solution 1 L HCl solution 1.50 mol HCl

15. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 15 Solution (continued) STEP 4 Set up problem to calculate needed quantity or concentration. 6.25 g Zn x 1.00 mol Zn x 1 mol H 2 = 0.0956 mol of H 2 65.41 g Zn 1 mol Zn 0.0200 L x 1.50 mol HCl x 1 mol H 2 = 0.0150 mol of H 2 1 L 2 mol HCl (smaller) Using the smaller number of moles of H 2.0150 mol H 2 x 22.4 L = 0.336 L of H 2 (B) 1 mol

16. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Summary of Calculations of Molarity and Chemical Reactions 16