Chapter4_1[1]

# Chapter4_1[1] - CHEMISTRY 1307 Chapter IV Stoichiometry...

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9/11/2011 1 1 CHEMISTRY 1307 Chapter IV Stoichiometry: Quantitative Information about Chemical Reactions 2 Stoichiometry The study of mass relationships in chemical reactions is called Stoichiometry . Stoichiometry provides quantitative information about chemical reactions. Mass must be ________ in a chemical reaction . Total mass of _________ Total mass of __________ = Chemical equations must therefore be balanced for mass. Numbers of atoms on the reactant side = Numbers of atoms on the product side 3 Stoichiometry The stoichiometric _________ are the numbers in front of the chemical formulas. They give the ratio of reactants and products. The balancing coefficients allow us to convert between numbers of reactants and products. ______ ___________ factors

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9/11/2011 2 4 Stoichiometry The ratio of any two species (reactants or products) in a balanced chemical reaction. Stoichiometric ratio: 2A + 3B A 2 B 3 2 A s combine with 3B s CONVERSION FACTORS!!! 2_ 3B 3B 2_ or 5 Stoichiometry _C 2 H 6 ( l ) + _ O 2 (g) _CO 2 (g) + _H 2 O( l ) Given the following reaction How many moles of water are produced when 3.0 mols of oxygen react? 2 2.6 mol H O 2 3.0 mol O 2 2 6 mol H O × 7 mol O mol O 2 mol H 2 O 2 sig. figs. 2 sig. figs. conversion factor 6 Stoichiometry Mole/Mass relationships. It is not possible to relate masses in reactions without going through _________!!!
9/11/2011 3 7 Stoichiometry If 454 g of solid ammonium nitrate decomposes to form dinitrogen monoxide gas and liquid water, how many grams of dinitrogen monoxide and water are formed? Data Information: Mass of reactant Write the balanced _______________ Step 1: Eq. gives mole ratios (stoichiometry) Convert ______ to moles Step 2: amount of reactant in moles Convert moles reactant to moles of _________ Step 3: amount of products in moles Convert moles of products to mass Step 4: mass of products in grams 8 Stoichiometry Step 1: Write the balanced chemical equation 43 1mol NH NO 454g NH NO × 80.04g NH NO = ____ mol NH 4 NO 3 Step 2: Convert mass of reactant (454 g NH 4 NO 3 ) to moles. NH 4 NO 3 (s) + _ H 2 O(l) N 2 O(g) Conversion (stoichiometric) factor 9 Stoichiometry Step 3: Using the balanced chemical equation, convert moles of reactant (NH 4 NO 3 ) to moles of products. 5.67 mol NH 4 NO 3 2 2 mol H O × 1 mol NH NO = ____ mol H 2 O Recall that the coefficients represent the molar ratios of reactants to products and vice versa… NH 4 NO 3 (s) + 2 H 2 O(g) N 2 O(g) 5.67 mol NH 4 NO 3 2 1 mol N O × 1 mol NH NO = 5.67 mol N 2 O

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9/11/2011 4 10 Stoichiometry Step 4: Now convert the moles of products (N 2 O and H 2 O) to grams using the molar mass of each.
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## This document was uploaded on 10/30/2011 for the course CHEM 1307 at Texas Tech.

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Chapter4_1[1] - CHEMISTRY 1307 Chapter IV Stoichiometry...

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