# 2S (s) + 3O2 (g) 2SO3 (g) O < 1

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2S (s) + 3O2 (g) 2SO3 (g) O2 3 0.03125 .666 < 1
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. The % yield in this experiment is __________. AND calculate the moles of unreacted excess reagent. Calculate the moles of the excess reagent that is unconsumed at the end ofreaction. MOLES = MASS/GFM S MOLES = 1.0 g s/ 32.04g/mol MOLES = mol S MOLES = MASS/GFM O2 MOLES = 1.0 g O2 / 32.00g/mol MOLES = mol O2 S = = , O .666 < The actual ratio is larger than the theoretical, therefore the actual ratio fraction is too large, the denominator is to small, O2 limiting.

INTRODUCING ICE CHART SOLUTION SKILL
THE ICE CHART WILL ALLOW YOU TO CALCULATE MOLES OF ALL SPECIES AFTER THE REACTION IS COMPLETED. MOLES OF ALL PRODUCTS. MOLES OF ALL EXCESS REACTANTS REMAINNING. THE FASTEST WAY TO APPROACH A REACTION IN WHICH YOU ARE CALCULATING MANY SUBSTANCES AT ONCE. 2S 3O  2SO3 INITIAL mol mol LIMITING 0.0 CHANGE -2X -3X END – 2X 3X

DETERMINING THE VALUE OF “X”
X IS DETERMINED BY THE LIMITING REAGENT. IN THIS PROBLEM mol = 3X FROM THE LIMITING REAGENT OXYGEN. X = MOL. THE EXCESS REAGENT IS SULFER, WHICH IS – 2X – 2( ) MOLE OF S REMAINS 4. THE PRODUCT SO3 = 2X = 2( ) = MOL ( GRAMS)

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