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Chapter 3 Formulas, Equations and Moles Lecture 6 (CHEM 1010) Agenda: Practice Quiz (Answer will be included in inked version) Problem 2.113, 2116 Limiting Reagents Problem Concentrations of Reactants in Solution: Molarity Diluting Concentrated Solutions Solution Stoichiometry Titration Percent Composition and Empirical Formulas Determining Empirical Formulas: Elemental Analysis Practice quiz The poisonous gas hydrogen cyanide (HCN) is produced by the high-temperature reaction of ammonia (NH 3 ) with methane (CH 4 ). Hydrogen is also produced in this reaction. (a)Write a balanced equation for the reaction that occurs. HCN produced (PRODUCT) from rxn NH 3 with CH 4 (REACTANTS) PS H 2 is also produced (PRODUCT) NH 3 + CH 4 HCN + H 2 N balance and C balance need to balance the Hs NH 3 + CH 4 HCN + 3H 2 Practice quiz The poisonous gas hydrogen cyanide (HCN) is produced by the high- temperature reaction of ammonia(NH 3 ) with methane (CH 4 ). Hydrogen is also produced in this reaction. (a) NH 3 + CH 4 HCN + 3H 2 (b) Suppose 500.0g of methane is mixed with 200.0 g of ammonia. MW(CH 4 ) = 12.01 + 4.04 = 16.05 g/mol Mol(CH 4 ) = 500.0 g / 16.05 g/mol = 31.15264 mol MW(NH 3 ) = 14.01 + 3.03 = 17.04 g/mol Mol(NH 3 ) = 200.0 g / 17.04 g/mol =11.73709 mol Limiting reagent is NH 3 Practice quiz Problem 2.113 5 th edition- The mass percent of an element in a compound is the mass of the element (total mass of the elements atoms in the molecule) divided by the mass of the compound (total mass of all atoms in the molecules) times 100%. What is the mass percent of each element in acetaminophen? C 8 H 9 NO 2 Problem 2.116 5 th edition Zinc has atomic mass A= 65.39 amu and has five naturally occurring isotopes: 64 Zn 63.929 amu 48.63% 66 Zn x amu 27.90% 67 Zn 66.927 amu 4.10% 68 Zn 67.925 amu 18.75% 70 Zn 69.925 amu 0.62% What is the isotopic mass of 66 Zn? Av atomic mass = (amu 64 Zn*% 64 Zn) +(amu 66 Zn*% 66 Zn) + . You know Av atomic mass and all the other numbers except for x. Solve for x. Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. g Al mol Al mol Fe 2 O 3 needed g Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al needed g Al needed 124 g Al 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 x = 367 g Fe 2 O 3 Start with 124 g Al need 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent. Use limiting reagent (Al) to calculate amount of product that can be formed.... View Full Document

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