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Chapter 12 Solutions

When a BaCl 2 solution is added to a Na 2 SO 4 solution, BaSO 4 , a white solid, forms. Chapter 12 Solutions. 12.6 Solutions in Chemical Reactions. Molarity in Chemical Reactions. In a chemical reaction ,

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Chapter 12 Solutions

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  1. When a BaCl2 solution is added to a Na2SO4 solution, BaSO4, a white solid, forms. Chapter 12 Solutions 12.6 Solutions in Chemical Reactions

  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. Calculations Involving Solutions in Chemical Reactions

  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 CaCO3? 2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l) STEP 1 State the given and needed quantities.Given 3.00 M HCl solution; 4.85 g of CaCO3 Need volume in milliliters STEP 2 Write a plan to calculate needed quantity or concentration. grams of CaCO3 moles of CaCO3 moles of HCl milliliters of HCl

  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 CaCO3 = 100.09 g of CaCO3 1 mol CaCO3 and 100.09 g CaCO3 100.09 g CaCO3 1 mol CaCO3 1 mol of CaCO3 = 2 mol of HCl 1 mol CaCO3 and 2 mol HCl 2 mol HCl 1 mol CaCO3 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. Example of Using Molarity in a Chemical Equation (continued) STEP 4 Set up problem to calculate needed quantity or concentration. 4.85 g CaCO3 x 1 mol CaCO3 x 2 mol HCl 100.09 g CaCO3 1 mol CaCO3 x 1000 mL HCl = 32.3 mL of HCl solution 3.00 mol HCl

  7. Learning Check How many milliliters of a 0.150 M Na2S solution are needed to react with 18.5 mL of a 0.225 M NiCl2 solution? NiCl2(aq) + Na2S(aq) NiS(s) + 2NaCl(aq) A. 4.16 mL B. 6.24 mL C. 27.8 mL

  8. Solution STEP 1 State the given and needed quantities.Given 0.0185 L of a 0.225 M NiCl2 solution; 0.150 M Na2S solution Need milliliters of Na2S solution STEP 2 Write a plan to calculate needed quantity or concentration. liters of NiCl2 solution moles of NiCl2 solution moles of Na2S solution milliliters of Na2S solution

  9. Solution (continued) STEP 3 Write equalities and conversion factors including mole-mole and concentration factors. 0.225 mol of NiCl2 = 1 L of NiCl2 solution 0.225 mol NiCl2 and 1 L NiCl2 1 L NiCl2 0.225 NiCl2 1 mol of NiCl2 = 1 mol of Na2S 1 mol NiCl2 and 1 mol Na2S 1 mol Na2S 1 mol NiCl2 1000 mL of Na2S solution = 0.150 mol of Na2S 1000 mL HCl and 0.150 mol HCl 0.150 mol HCl 1000 mL HCl

  10. Solution (continued) STEP 4 Set up problem to calculate needed quantity or concentration. 0.0185 L x 0.225 mol NiCl2x 1 mol Na2S x 1000 mL 1 L 1 mol NiCl2 0.150 mol = 27.8 mL of Na2S solution (C)

  11. Learning Check How many liters of H2 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) ZnCl2 (aq) + H2(g) A. 4.28 L of H2 B. 0.336 L of H2 C. 0.168 L of H2

  12. 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 H2 gas at STP STEP 2 Write a plan to calculate needed quantity or concentration. Limiting reactant: lowest number of moles of H2 1) grams of zinc moles of zinc moles of H2 2) liters of HCl solution moles of HCl solution moles of H2

  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 H2 1 mol Zn and 1 mol H2 1 mol H2 1 mol Zn 2 mol of HCl = 1 mol of H2 2 mol HCl and 1 mol H2 1 mol H2 2 mol HCl

  14. Solution (continued) STEP 3 (continued) 1 mol of H2 = 22.4 L of H2 22.4 L H2 and 1 mol H2 1 mol H2 22.L H2 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. Solution (continued) STEP 4Set up problem to calculate needed quantity or concentration. 6.25 g Zn x 1.00 mol Zn x 1 mol H2 = 0.0956 mol of H2 65.41 g Zn 1 mol Zn 0.0200 L x 1.50 mol HCl x 1 mol H2 = 0.0150 mol of H2 1 L 2 mol HCl (smaller) Using the smaller number of moles of H2 1.50 mol H2 x 22.4 L = 0.336 L of H2(B) 1 mol

  16. Summary of Calculations of Molarity and Chemical Reactions

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