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Ch_3-4_Limiting_reactant_and_yield

Ch_3-4_Limiting_reactant_and_yield - LIMITING REACTANT...

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EXAMPLE: The reaction of Cl 2 with P 4 produces PCl 5 . If 6.00 g of chlorine is mixed with 2.20 g phosphorous, what is the maximum yield? LIMITING REACTANT Molar masses: Cl 2 2 x 35.45 = 70.90 70.9 g/mol P 4 4 x 30.97 = 123.88 124 g/mol PCl 5 30.97 + 5 x 35.45 = 208.22 208 g/mol P 4 + Cl 2 PCl 5 P 4 + 10Cl 2 4PCl 5 Balanced equation 2.20 g 6.00 g Calculate yield using 2.20 g P 4 and assuming excess Cl 2 : 5 5 4 4 5 4 4 5 4 208 2.20 1 14.8 1 124 1 1 mol PCl g PCl g P mol P x x x g PCl g P mol P mol PCl = Unbalanced equation Calculate yield using 6.00 g Cl 2 and assuming excess P 4 : 5 5 2 2 5 2 2 5 4 208 6.00 1 7.04 1 70.9 10 1 mol PCl g PCl g Cl mol Cl x x x g PCl g Cl mol Cl mol PCl = Correct answer is smaller value.
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Why smaller value is the correct answer: Calculate yield using 2.20 g P 4 and assuming excess Cl 2 : 5 5 4 4 5 4 4 5 4 208 2.20 1 14.8 1 124 1 1 mol PCl g PCl g P mol P x x x g PCl g P mol P mol PCl = How much chlorine is required to react with 2.20 g phosphorus? P 4 + 10Cl 2 4PCl 5 4 4 2 2 2 4 4 2 2.20 1 10 70.9 12.6 1 124 1 1 g P mol P mol Cl g Cl x x x g Cl g P mol P mol Cl = Since there are only 6.00 g Cl
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