Fund. M - Limiting Reagents & Combustion Analysis

Fund. M - Limiting Reagents & Combustion Analysis -...

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Fundamentals M – Limiting Reactants M.2 – The Limits of Reaction If we have 400 g SO 2 , 175 g O 2 & 125 g H 2 O – what is the limiting reagent? SO 2 + O 2 + H 2 O → H 2 SO 4 Balance eq., 2 SO 2 + O 2 + 2 H 2 O → 2 H 2 SO 4 Convert all mass to moles, then use mole to mole ratio to find amt of product Moles SO 2 = 6.24 moles x 2 mol H 2 SO 4 = 6.24 mol H 2 SO 4 2 mol SO 2 Moles O 2 = 5.47 moles x 2 mol H 2 SO 4 = 10.94 mol H 2 SO 4 1 mol O 2 Moles H 2 O = 6.94 moles x 2 mol H 2 SO 4 = 6.94 mol H 2 SO 4 1 mol H 2 O The limiting reagent is the one that make the least amount of product: SO 2 is the limiting reagent! O 2 & H 2 O are present in excess. How much H 2 SO 4 will be produced? 6.24 mol H 2 SO 4 x 98.07 g = 612 g of H 2 SO 4 will be produced 1 mol How much O 2 will be consumed? 6.24 moles SO 2 x 1 mol O 2 = 3.12 moles of O 2 will be consumed in the rxn 2 mol SO 2 How much O 2 will remain? 5.47 mol O 2 3.12 mol O 2 = 2.35 moles of O 2 will remain 2.35 moles x 32.00 g = 75.2 g of O 2 will remain 1 mol M.1 – Reaction Yield If my actual yield of the above reaction was 584 g of H 2 SO 4 what is the percent yield? Percent yield = actual yield x 100 584 x 100 = 95.4% theoretical yield 612
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M.3 – Combustion Analysis A certain compound, used as a welding fluid, contains carbon & hydrogen only. Burning a sample of it completely in oxygen gives 3.38 g of CO2, 0.0692 g of water and no other products. What is the empirical formula of this compound? Need to know that combustion is with O 2 as a reactant and CO 2 & H 2 O as products C x H x + O 2 → CO 2 and H 2 O Because all the carbon has been converted to CO 2 and all the hydrogen has been converted to H 2 O we can determine the amounts in the reactant. 3.38 g CO 2 = 1 mole = 0.0768 mol of CO 2 x 1 mol C = 0.0768 mol C 44.01 g 1 mol CO 2 0.692 g H 2 O = 1 mole = 0.0384 mol of H 2 O x 2 mol H = 0.0768 mol H 18.02 g 1 mol H 2 O 0.00768 = 1 empirical formula = CH 0.00768 Find molecular formula: the molar mass of this welding fluid is 25.9 g/mol Calculate the empirical formula molar mass CH = 13.0 g/mol 25.9 = 1.99 so multiple 2(CH) = C 2 H 2 13.0
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Fundamentals M – Limiting Reactants Questions 1.
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This note was uploaded on 04/24/2008 for the course CHEM 6A taught by Professor Pomeroy during the Winter '08 term at UCSD.

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Fund. M - Limiting Reagents & Combustion Analysis -...

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