Ch7 - 0.000 3.25 59.2 PROCEDURE#2 † How to Solve a...

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PROCEDURE #1 † How to Solve a Limiting Reactant problem .---optional method 5b Comparison-of-moles of Reactants Method. Convert the number of grams of each reactant to moles. Divide the number of grams of each by its molecular weight. Identify the limiting reactant. Calculate the moles needed and compare to the moles available for one reactant compared to the other one. Calculate the number of moles of each species that reacts or is produced. Calculate the number of moles of each species that remains after the reaction. Change the number of moles of each species to grams. Summary table: 2Fe(s) + 3Cl 2 (g) 2FeCl 3 (s) Fe Cl 2 FeCl 3 Grams at start 20.4 42.1 0 Molar mass, g/mol 55.85 70.90 162.2 Moles at start 0.365 0.594 0 Change in moles - 0.365 - 0.548 + 0.365 Moles at end 0.000 4.59E-2 0.365 Grams at end
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Unformatted text preview: 0.000 3.25 59.2 PROCEDURE#2 † How to Solve a Limiting Reactant problem. .---required method 6 Smaller Amount of Product Method. Calculate the amount of product that can be formed by the initial amount of each reactant. (A) The reactant that gives the smaller amount of product is the limiting reactant. (B) The smaller amount of product is the amount that will be formed when the limiting reactant is used up. Calculate the amount of the non-limiting reactant that is needed to use up the limiting reactant. Subtract the amount of non-limiting reactant needed to use up the limiting reactant from the original amount of non-limiting reactant. The difference is the excess amount of the non-limiting reactant....
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This note was uploaded on 04/10/2008 for the course CHEM 135 taught by Professor Brooks during the Spring '08 term at Maryland.

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