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Basic Stoichiometry Reactions to Calculate Molecule Counts

Gain insight into stoichiometry, the method of quantifying reactants and products in reactions. Dive into examples and learn to identify limiting and excess reactants.

amery-hendricks
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Basic Stoichiometry Reactions to Calculate Molecule Counts

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  1. Reactions- Stoichiometry Things I added are in RED (info in parentheses are explanations)

  2. How many molecules do we need for each equation below? • Na+ + Cl- NaCl Na: 1, Cl: 1  NaCl: 1 we have 1 Na,1 Cl, and the react to from 1 NaCl • 2Mg + O2  2MgOwe need to balance the eqn Mg: 2, O2*: 1  MgO2 molecules of Mg, 1 MOLECULE OF O2* will yeild 2 molecules of MgO • 2KOH + MgSO4  Mg(OH)2 + K2SO4 KOH: 2*, MgSO4: 1*  Mg(OH)2: 1* + K2SO4: 1* It’s important to think about things like this: if we have 2 molecules of KOH, we can make 1 molecule of Mg(OH)2; if we have 6 molecules of KOH, we can make 3 Mg(OH)2 *In early classes, we looked at ATOMs instead of molecules. Looking at molecules makes this easier; these notes reflect that change.

  3. Stoichiometry • Deals with the amounts of atoms or molecules involved in a reaction. This is a basic definition we’ll use to describe stoichiometry

  4. Questions • What if we have the following: • 100 Na+ + 10 Cl- 10 NaCl + 90 Na+ • Na+ will the the Excess reactant; Cl- is the limiting reactant • Number of ions • reactants: Na: 100, Cl: 10 • products: Cl: 10 (all from NaCl); Na: 10 (from NaCl) + 90 Na (leftovers) = 100 • 100 Mg + 10 O2  20 MgO + 80 Mg • Mg is the excess reactant; O2 is the limiting reactant • Since it takes 1 O2 to make 2 MgO, we end up w/ 20 MgO (see slide 2 for balanced eqn) • Number of molecules • Reactants: Mg: 100; O2: 10 OR 20 atoms of O • Products: Mg: 20 (MgO) + 80 (Mg) = 100; O2: 10 (20 MgO) OR O: 20 (20 MgO)

  5. Terms • Limiting reactant: The reactant that is used up first; it LIMITS the amount of product that can be made • It not always the reactant that has the smallest number in front of it, though students often think this. See the next slide, first problem for this. • Excess reactant: The reactant that remains after the limiting reactant is used up

  6. Which is the limiting reactant? • CuCl2 + 2Li  Cu + 2LiCl (balanced equation) • 12 CuCl2 + 21 Li  • 12 CuCl2 x (2 Li/ 1 CuCl2)(from the balanced eqn)= 24 Li CuCl2/CuCl2(the CuCl2 cancel out); 24 Li are needed to fully react the 12 CuCl2 Li is the limiting reactant b/c they are only 21 < 24 needed • 2N2 + 5O2 2N2O5 (balanced equation) • 10 N2 + 30 O2 10 N2 x (5 O2/ 2 N2) = 25 O2 N2/N2; 25 O2 are needed to fully react the 10 N2 Since 30 > 25, N2 is the limiting reactant; O2 is the excess reactant

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