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Unformatted text preview: ? 2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g) 1 mo l N2H4 3 mo l N2 100 g N2H4 x ¾¾¾¾¾ x ¾¾¾¾¾ = 4.68 mo l N2 32.04 g N2H4 2 mo l N2H4 Limiting Reactant 1 mo l N2O4 3 mo l N2 200 g N2O4 x ¾¾¾¾¾ x ¾¾¾¾¾ = 6.52 mol N2 92.00 g N2O4 1 mo l N2O4 28.0 g N2 4.68 mo l N2 x ¾¾¾¾¾ = 131 g N2 1 mo l N2 Notes: 1. Even though mass of N2O4 is greater than N2H4, there are fewer mo les of it due to its larger mo lar mass. 2. Limit ing reactant calculations must always be done when amount of both reactants are given. 26 Chemistry 51 Chapter 6 LIMITING REACTANT Examples: 3. How many mo les of Fe3O4 can be produced by reacting 16.8 g of Fe with 10.0 g of H2O as shown below: 3 Fe (s) + 4 H2O (g) D ¾¾ Fe3O4 (s) + 4 H2 (g) ® Assume Fe is the limit ing reactant: 16.8 g Fe x ¾¾¾¾¾¾¾¾ x ¾¾¾¾¾¾¾ = mo l Fe3O4 Assume H2O is the limit ing reactant: 10.0 g H2O x ¾¾¾¾¾¾¾¾ x ¾¾¾¾¾¾¾ = mo l Fe3O4 Correct limit ing reactant is _________ and __________ mo l of Fe3O4 are produced. 4. How many grams o f AgBr can be produced when 50.0 g of MgBr2 is mixed with 100.0 g of AgNO3, as shown below: MgBr2 (aq) + 2 AgNO3 (aq) ¾¾ 2 AgBr (s) + Mg(NO3)2 (aq) ® 27 Chemistry 51 Chapter 6 PERCENT YIELD · The amount of product calculated through stoichiometric ratios is the maximum amount of product that can be produced during the reaction, and is thus called theoretical yield. · The actual yield of a product in a chemical react ion is the actual amount obtained fro m the reaction. · The percent yield of a reaction is obtained as fo llows: Actual yield x100 = Percent yield Theoretical yield Example: 1. In an experiment forming ethano l, the theoretical yield is 50.5 g and the...
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This document was uploaded on 03/18/2014 for the course CHEM 51 at Los Angeles Mission College.

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