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Reactions- Stoichiometry

Reactions- Stoichiometry. Things I added are in RED (info in parentheses are explanations). How many molecules do we need for each equation below?. Na + + Cl -  Na Cl Na: 1, Cl : 1  NaCl : 1 we have 1 Na, 1 Cl , and the react to from 1 NaCl

amery-hendricks
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Reactions- Stoichiometry

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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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