Chapter 15 Study Guide

Chapter 15 Study Guide - limiting reactant given. Molarity...

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Chapter 15 Test – STUDY GUIDE Stoichiometry Molar Ratio Moles of reactant x (coefficient of product / coefficient of reactant) = moles of products - note: you can rearrange this like an algebraic equation Limiting Reactant Remember: ONE of your reactants will always run out. 1. Determine the molar ratio for the reactants, label them (in order) R1 and R2. 2. Solve for the number of moles of R2 needed to use up ALL of R1 a. Moles of R1 x (coef of R2 / coef of R1) = Moles of R2 needed 3. If the number of moles of R2 needed is less than the moles of R2 given, then R1 is your limiting reactant. If not, you need to repeat step 2 swapping R1 with R2 and R2 will be your limiting reactant. 4. Solve for the amount of excess reactant left over. a. Moles of E.R. given – moles of E.R. needed = moles of E.R. leftover 5. Solve for the amount of product produced using stoichiometry and the amount of
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Unformatted text preview: limiting reactant given. Molarity Molarity = moles / liter or M = n / V Dilutions Moles will be equal for solutions 1 and 2. Solution 1 is the solution you have, and solution 2 is the solution you are making. (soln = solution) Step 1: Find the number of moles using soln 2. M 2 x V 2 = n Step 2: The number of moles for both solns is equal. Step 3: Find the volume of soln 1 using the moles and M of soln 1 given. n / M 1 = V 1 Step 4 (optional): Find the volume of water to add by subtracting. V 2 V 1 = V water: Saturation Solvent runs out of space to store solute Under Saturation- capable of dissolving more solute Supper Saturation- contains more of dissolved solute Solution Homogeneous mix of solute and solvent Solute- smaller quantity; chemical that is being dissolved Solvent- what dissolves the solute...
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Chapter 15 Study Guide - limiting reactant given. Molarity...

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