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Chemistry SI Review: Limiting Reactants, Empirical Formulas, and Ionic Equations, Study notes of Chemistry

A chemistry si review covering topics such as determining the limiting reactant, writing empirical and molecular formulas, and writing net ionic and ionic equations. The review includes examples and calculations for each topic.

Typology: Study notes

Pre 2010

Uploaded on 08/31/2009

koofers-user-eyw
koofers-user-eyw 🇺🇸

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Download Chemistry SI Review: Limiting Reactants, Empirical Formulas, and Ionic Equations and more Study notes Chemistry in PDF only on Docsity! SI for Dr. Rhyne’s Chem 1101—SI leader Hilary Gantt Chemistry Chapter 4 review KEY 1) Consider the reaction 2 Al(s) + 3 Cl2(g) à 2AlCl3(s). A mixture of 1.50 mol of AL & 3.00 mol of Cl2 are allowed to react. 1.50 mol Al | 2 mol ALCl3 = 1.50 AlCl3 | 2 mol Al 3.00 mol Cl2 | 2 mol AlCl3 = 2.00 mol AlCl3 | 3 mol Cl2 a) What’s the Limiting Reactant? Al b) How many mol of AlCl3 are formed? 1.50 mol c) How many moles of the excess reactant remain at the end of the reaction? 0.75 mol Cl2 2.00-1.50 = 0.50 mol AlCl3 0.50 mol AlCl3 | 3 mol Cl2 = 0.75 mol Cl2 | 2 mol AlCl3 2) Determine the empirical and molecular formulas for each of the following species: caffeine- 49.5% C, 5.15% H, 28.9% N, 16.5% by mass; molar mass=195 g/mol 49.5 g C | 1 mol C = 4.12 mol C/1.03= 4 | 12.01 g C 5.15 g H | 1 mol H = 5.11 mol H/1.03 = 5 | 1.008 g H 28.9 g N | 1 mol N = 2.06 mol N/1.03 = 2 | 14.01 g N 16.5 g O | 1 mol O = 1.03 mol O/ 1.03 = 1 | 16.00 g O C4H5N2O FW = 97 195/97=2 2(C4H5N2O) = C8H10N4O2
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