#5 of 11-6 LIMITING REAGENT SET

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#5 of 11-6 LIMITING REAGENT SET INTRODUCTION OF ICE BOX STOICHIOMETRY SOLUTION TABLES. ICE BOX SOLUTION TABLES: ARE THE MOST EFFICIENT TO SOLVE FOR EVERYTHING WHAT IS EVERYTHING? REAGENTS LEFT OVER (UNCONSUMED) PRODUCT YIELD REACTANTS CONSUMED TO 0.0 MOLES. 5) Sulfur and oxygen react in a combination reaction to produce sulfur trioxide, an environmental 5) pollutant: 2S (s) + 3O2 (g)  2SO3 (g) In a particular experiment, the reaction of 1.0 g S with 1.0 g O2 produced 0.80 g of SO3. Calculate the amount of excess reagent that remains not having reacted (in excess).

Defining X X is defined by the LIMMITING REAGENT. Divide the moles of the limiting reagent by the coefficient of the balanced reaction. In this example oxygen is the limiting reagent, therefore 3 X = 0.03125 mol X = 0.01041666 mole 2S + 3O2  2SO3 0.031191 mol 0.03125 mol 0.0 -2 X -3 X +2 X 0.031191 mol -2x 2 (.01041666 ) = 0.020833 0.031191 mol -2x : 0.031191 mol - 2 (0.01041666 mol) 0.031191 mol -(0.020833) = 01041666 mol S remains