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Chemistry Notes: Reactions in Solutions

Chemistry Notes: Reactions in Solutions Two solutions can be combined to generate a chemical reaction. Often, chemical reactions occur in aqueous * solutions, e.g. living systems.

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Chemistry Notes: Reactions in Solutions

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  1. Chemistry Notes: Reactions in Solutions • Two solutions can be combined to generate a chemical reaction. • Often, chemical reactions occur in aqueous* solutions, e.g. living systems. • The concentration of each solution and the types of solutes involved must be taken into account. * Water = solvent • ------------------------------------------------------------------------------------- • Example of a chemical reaction: • HCl + NaOH --> NaCl + H2O. • Let’s suppose this reaction occurs as a result of combining two solutions: HCl & NaOH. • Let: HCl solution = 1.0 M (= 1.0 mol HCl/L) • NaOH solution = 1.0 M (= 1.0 mol NaOH/L) • Combine 10 mL of each solution together. How many grams of NaCl are produced? • 1st: Determine how many moles of each reactant are used. • 10 mL of 1.0 M HCl = 10 mL x 1.0 mol = 0.01 mol HCl • 1000 L • 10 mL of 1.0 M HCl = 10 mL x 1.0 mol = 0.01 mol NaOH • 1000 L Divide mL by 1000 to convert to L.

  2. Next, determine moles of product produced (stoichiometry). HCl + NaOH --> NaCl + H2O 0.01 + 0.01 --> 0.01 0.01 mol of NaCl produced. mol molmol Then, convert moles to grams. (0.01 mol)(23+35)g/mol = 0.58 g NaCl produced Molar mass NaCl

  3. Example 2: • 15.0 mL of a 0.25 M solution of AgNO3 is combined with 10.0 mL of a 0.35 M solution of KBr in the following reaction: • AgNO3 + KBr ---> AgBr + KNO3 (balanced) • 1) What is the limiting reagent? • 2) How many grams of AgBr are produced? • Step 1: Determine the moles of each reactant used. • AgNO3 = (15 mL/1000)(0.25 mol/L) = 0.00375 mol AgNO3 • KBr = (10 mL/1000)(0.35 mol/L) = 0.0035 mol KBr • Step 2: Determine limiting reagent. • AgNO3 + KBr ---> AgBr + KNO3 • Molar ratios are all 1, so the limiting reagent will be determined by the smaller quantity of reactant, which is KBr @ 0.0035 mol. • Step 3: Determine the moles of product. • 0.0035 mol KBr ---> 0.0035 mol AgBr • Step 4: Calculate the grams of AgBr. • (0.0035 mol AgBr)(108+80)g/mol = 0.66 gAgBr produced

  4. You do one. • 50 mL of a 5.00 M solution of Li2SO4 is combined with 45 mL of a 4.50 M solution of MgCl2, according to the following reaction: • Li2SO4 + MgCl2 ---> 2LiCl + MgSO4. •  How many grams of MgSO4 are produced? • Remember: • Step 1: Detemine the moles pf each reactant. • Step 2: Determine the limiting reagent. • Step 3: Determine the moles of the product. • Step 4: Calculate the grams of the product. Molar mass AgBr

  5. Solution Step 1. (50 mL/1000)(5.00 M) = 0.2500 mol Li2SO4 (45 mL/1000)(4.50 M) = 0.2025 mol MgCl2 Step 2. Molar ratios for both reactants as well as the product are 1:1 so the smaller quantity of reactant (mol) = the quantity of product. Smaller quantity of reactant = MgCl2 @ 0.2025 mol … Step 3. … ---> 0.2025 mol MgSO4. Step 4. (0.2025 mol MgSO4)(24+32+16x4) = 24.3 g MgSO4.

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