Presentation is loading. Please wait.

Presentation is loading. Please wait.

Sample Problem 3.10Calculating Amounts of Reactant and Product in a Limiting- Reactant Problem PROBLEM: A fuel mixture used in the early days of rocketry.

Similar presentations


Presentation on theme: "Sample Problem 3.10Calculating 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 Problem PROBLEM: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N 2 H 4 ) and dinitrogen tetraoxide(N 2 O 4 ), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x10 2 g of N 2 H 4 and 2.00x10 2 g of N 2 O 4 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. mol of N 2 divide by M molar ratio mass of N 2 H 4 mol of N 2 H 4 mass of N 2 O 4 mol of N 2 O 4 limiting mol N 2 g N 2 multiply by M

2 Sample Problem 3.10Calculating Amounts of Reactant and Product in a Limiting- Reactant Problem continued SOLUTION: N 2 H 4 ( l ) + N 2 O 4 ( l ) N 2 ( g ) + H 2 O( l ) 1.00x10 2 g N 2 H 4 = 3.12mol N 2 H 4 mol N 2 H g N 2 H mol N 2 H 4 = 4.68mol N 2 3 mol N 2 2mol N 2 H x10 2 g N 2 O 4 = 2.17mol N 2 O 4 mol N 2 O g N 2 O mol N 2 O 4 = 6.51mol N 2 3 mol N 2 mol N 2 O 4 N 2 H 4 is the limiting reactant because it produces less product, N 2, than does N 2 O mol N 2 mol N g N 2 = 131g N 2 243

3 A + B (reactants) C (main product) D (side products) The effect of side reactions on yield. Figure 3.9

4 Sample Problem 3.11 Calculating Percent Yield PROBLEM: Silicon carbide (SiC) is an important ceramic material that is made by allowing sand (silicon dioxide, SiO 2 ) 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: write balanced equation find mol reactant & product find g product predicted percent yield actual yield/theoretical yield x 100 SOLUTION: SiO 2 ( s ) + 3C( s ) SiC( s ) + 2CO( g ) kg SiO 2 mol SiO g SiO g SiO 2 kg SiO 2 = 1664 mol SiO 2 mol SiO 2 = mol SiC = mol SiC g SiC mol SiC kg 10 3 g = kg x 100=77.0% 51.4 kg kg

5 5 1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units Amounts of Reactants and Products

6 6 Limiting Reagent: 2NO + O 2 2NO 2 NO is the limiting reagent O 2 is the excess reagent Reactant used up first in the reaction.

7 7 In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2 Al + Fe 2 O 3 Al 2 O Fe Calculate the mass of Al 2 O 3 formed. g Almol Almol Fe 2 O 3 neededg Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al neededg Al needed 124 g Al 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 x = 367 g Fe 2 O 3 Start with 124 g Alneed 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent

8 8 Use limiting reagent (Al) to calculate amount of product that can be formed. g Almol Almol Al 2 O 3 g Al 2 O g Al 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al x 102. g Al 2 O 3 1 mol Al 2 O 3 x = 234 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe At this point, all the Al is consumed and Fe 2 O 3 remains in excess.

9 9 Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100% Reaction Yield


Download ppt "Sample Problem 3.10Calculating Amounts of Reactant and Product in a Limiting- Reactant Problem PROBLEM: A fuel mixture used in the early days of rocketry."

Similar presentations


Ads by Google