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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 + H 2 O. 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? 1 st : 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.

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

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Example 2: 15.0 mL of a 0.25 M solution of AgNO 3 is combined with 10.0 mL of a 0.35 M solution of KBr in the following reaction: AgNO 3 + KBr ---> AgBr + KNO 3 (balanced) 1) What is the limiting reagent? 2) How many grams of AgBr are produced? Step 1: Determine the moles of each reactant used. AgNO 3 = (15 mL/1000)(0.25 mol / L ) = 0.00375 mol AgNO 3 KBr = (10 mL/1000)(0.35 mol / L ) = 0.0035 mol KBr Step 2: Determine limiting reagent. AgNO 3 + KBr ---> AgBr + KNO 3 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 g AgBr produced

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You do one. 50 mL of a 5.00 M solution of Li 2 SO 4 is combined with 45 mL of a 4.50 M solution of MgCl 2, according to the following reaction: Li 2 SO 4 + MgCl 2 ---> 2LiCl + MgSO 4. 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

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Solution Step 1.(50 mL/1000)(5.00 M) = 0.2500 mol Li 2 SO 4 (45 mL/1000)(4.50 M) = 0.2025 mol MgCl 2 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 = MgCl 2 @ 0.2025 mol … Step 3. … ---> 0.2025 mol MgSO 4. Step 4. (0.2025 mol MgSO 4 )(24+32+16x4) = 24.3 g MgSO 4.

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