Presentation on theme: "Sample Problem 3.10 Calculating Amounts of Reactant and Product in a Limiting-Reactant Problem PROBLEM: A fuel mixture used in the early days of rocketry."— Presentation transcript:
1 Sample Problem 3.10Calculating Amounts of Reactant and Product in a Limiting-Reactant ProblemPROBLEM:A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x102 g of N2H4 and 2.00x102 g of N2O4 are mixed?PLAN:We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given.In this case one of the reactants is in molar excess and the other will limit the extent of the reaction.mass of N2H4mass of N2O4limiting mol N2divide by Mmultiply by Mmol of N2H4mol of N2O4g N2molar ratiomol of N2mol of N2
2 Sample Problem 3.10Calculating Amounts of Reactant and Product in a Limiting-Reactant ProblemcontinuedSOLUTION:2N2H4(l) + N2O4(l) N2(g) H2O(l)34mol N2H432.05g N2H41.00x102g N2H4= 3.12mol N2H4N2H4 is the limiting reactant because it produces less product, N2, than does N2O4.3 mol N22mol N2H43.12mol N2H4= 4.68mol N2mol N228.02g N24.68mol N2= 131g N2mol N2O492.02g N2O42.00x102g N2O4= 2.17mol N2O43 mol N2mol N2O42.17mol N2O4= 6.51mol N2
3 C A + B D The effect of side reactions on yield. Figure 3.9 (main product)A + B(reactants)D(side products)
4 Sample Problem 3.11Calculating Percent YieldPROBLEM:Silicon carbide (SiC) is an important ceramic material that is made by allowing sand (silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When kg of sand is processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC from this process?PLAN:SOLUTION:SiO2(s) + 3C(s) SiC(s) + 2CO(g)write balanced equation103 g SiO2kg SiO2mol SiO260.09 g SiO2100.0 kg SiO2= 1664 mol SiO2find mol reactant & productmol SiO2 = mol SiC = 1664find g product predicted40.10 g SiCmol SiCkg103g1664 mol SiC= kgactual yield/theoretical yield x 10051.4 kg66.73 kgpercent yieldx 100=77.0%
5 Amounts of Reactants and Products Write balanced chemical equationConvert quantities of known substances into molesUse coefficients in balanced equation to calculate the number of moles of the sought quantityConvert moles of sought quantity into desired units
6 Limiting Reagent: Reactant used up first in the reaction. 2NO + O NO2NO is the limiting reagentO2 is the excess reagent
7 Have more Fe2O3 (601 g) so Al is limiting reagent In one process, 124 g of Al are reacted with 601 g of Fe2O32 Al + Fe2O Al2O3 + 2 FeCalculate the mass of Al2O3 formed.g Almol Almol Fe2O3 neededg Fe2O3 neededORg Fe2O3mol Fe2O3mol Al neededg Al needed1 mol Al27.0 g Alx1 mol Fe2O32 mol Alx160. g Fe2O31 mol Fe2O3x=124 g Al367 g Fe2O3Start with 124 g Alneed 367 g Fe2O3Have more Fe2O3 (601 g) so Al is limiting reagent
8 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al x 102. g Al2O3 1 mol Al2O3 x Use limiting reagent (Al) to calculate amount of product thatcan be formed.g Almol Almol Al2O3g Al2O32Al + Fe2O Al2O3 + 2Fe1 mol Al27.0 g Alx1 mol Al2O32 mol Alx102. g Al2O31 mol Al2O3x=124 g Al234 g Al2O3At this point, all the Al is consumed and Fe2O3 remains in excess.
9 Reaction Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted.Actual Yield is the amount of product actually obtainedfrom a reaction.% Yield =Actual YieldTheoretical Yieldx 100%
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