The Combustion Reaction Of C3H8O Is Represented By Class 11 ...

CoursesCourses for KidsFree study materialOffline CentresMoreStore

Talk to our experts

1800-120-456-456

Sign In

• Question Answer

• Class 11

• Chemistry

• The combustion reaction of C3H...

AnswerQuestion Answers for Class 12Class 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11Class 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10Class 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9Class 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8Class 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7Class 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6Class 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5Class 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4Class 4 ScienceClass 4 EnglishClass 4 MathsThe combustion reaction of ${C_3}{H_8}O$ is represented by this balanced chemical equation: $2{C_3}{H_8}O + 9{O_2} \to 6C{O_2} + 8{H_2}O$. When $6.01 \times {10^{22}}$ molecules of ${O_2}$ and $2.78 \times {10^{23}}$ molecules of ${C_3}{H_8}O$ are combined, how many grams of which reagent is used?AnswerVerified495.3k+ viewsHint: Some chemical reactions contain more than one reactant which are present in different quantities. As the reaction proceeds consumption of any reagent will cause a stoppage of reaction. The reagent which is exhausted first is known as the limiting reagent.Complete answer:Organic molecules undergo the process of combustion to form carbon dioxide and water molecules. ${C_3}{H_8}O$ chemically is propanol which undergoes combustion with nine moles of oxygen molecule ${O_2}$to form six mole of $C{O_2}$ and eight moles of ${H_2}O$.From the balanced chemical reaction we see that for every one mole of ${C_3}{H_8}O$ nine moles of ${O_2}$ is required. Hence, the ratio of these two reagents is $2:9$.It means that if $2.78 \times {10^{23}}$ molecules of ${C_3}{H_8}O$ is used then according to ratio $8.28 \times {10^{25}}$ molecules of ${O_2}$ must be required.As from the question we see that we have a smaller number of oxygen molecules than what we require for complete reaction with ${C_3}{H_8}O$. Hence, oxygen is exhausted first and acts as a limiting reagent for the chemical reaction. So, we have to calculate the value of all reagents with respect to oxygen.We have $6.01 \times {10^{22}}$ molecules of ${O_2}$ which when undergoing reaction require ${C_3}{H_8}O$ in a ratio of $2:9$.Hence when all the $6.01 \times {10^{22}}$ are used, $1.336 \times {10^{22}}$ molecules of ${C_3}{H_8}O$ are required for complete reaction.Hence, we see that some molecules of ${C_3}{H_8}O$ remain unreacted during the chemical reaction-Total molecules of ${C_3}{H_8}O$ remaining after the reaction $ = $ total molecules of ${C_3}{H_8}O$$ - $ molecules used during the chemical reactionTotal molecules of ${C_3}{H_8}O$ remaining after the reaction $ = $$2.78 \times {10^{23}}$$ - $$1.336 \times {10^{22}}$Total molecules of ${C_3}{H_8}O$ remaining after the reaction$ = $$2.642 \times {10^{23}}$Mass of particle equal to Avogadro number is equal to atomic mass or molecular mass of the compound For oxygen:$6.022 \times {10^{24}}$ molecules of ${O_2}$ $ = $ $32g$${O_2}$Hence one weight of one molecule of ${O_2}$ is$1$ molecules of ${O_2}$ $ = $ $\dfrac{{32}}{{6.022 \times {{10}^{24}}}}g$${O_2}$$6.01 \times {10^{22}}$ molecules of ${O_2}$ $ = $ $\dfrac{{6.01 \times {{10}^{22}}}}{{6.022 \times {{10}^{24}}}} \times 32g$${O_2}$After the calculation, we find that $6.01 \times {10^{22}}$ molecules of ${O_2}$$ = $ $0.319g$${O_2}$Hence, $0.319g$${O_2}$ is required in the combustion process.For ${C_3}{H_8}O$:$6.022 \times {10^{24}}$ molecules of ${C_3}{H_8}O$$ = $ $60g$${C_3}{H_8}O$Hence one weight of one molecule of ${C_3}{H_8}O$ is$1$ molecules of ${C_3}{H_8}O$ $ = $ $\dfrac{{60}}{{6.022 \times {{10}^{24}}}}g$${C_3}{H_8}O$We already discussed that $1.336 \times {10^{22}}$ molecules of ${C_3}{H_8}O$ are utilized during the reaction:$1.336 \times {10^{22}}$ molecules of ${C_3}{H_8}O$ $ = $ $\dfrac{{1.336 \times {{10}^{22}}}}{{6.022 \times {{10}^{24}}}} \times 60g$${C_3}{H_8}O$After the calculation, we find that $1.336 \times {10^{22}}$ molecules of ${C_3}{H_8}O$$ = $ $0.132g$${C_3}{H_8}O$$ \Rightarrow $ After calculation we find that for the combustion reaction $2{C_3}{H_8}O + 9{O_2} \to 6C{O_2} + 8{H_2}O$, $0.319g$${O_2}$ and $0.132g$${C_3}{H_8}O$ are required.Note:In chemistry stoichiometry is used to relate quantities of different reactants and products in a chemical reaction. In a chemical reaction whole numerical digits are used to write the balanced chemical equation. Stoichiometry helps in identifying the concentration of the component from the chemical equation.Recently Updated PagesMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Economics: Engaging Questions & Answers for SuccessMaster Class 11 English: Engaging Questions & Answers for SuccessMaster Class 11 Maths: Engaging Questions & Answers for SuccessMaster Class 11 Biology: Engaging Questions & Answers for SuccessMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Economics: Engaging Questions & Answers for SuccessMaster Class 11 English: Engaging Questions & Answers for SuccessMaster Class 11 Maths: Engaging Questions & Answers for SuccessMaster Class 11 Biology: Engaging Questions & Answers for Success

• 1
• 2

Trending doubtsThere are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSEOne Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSEExplain zero factorial class 11 maths CBSEState the laws of reflection of light10 examples of friction in our daily lifeWho is known as the father of chemistry class 11 chemistry CBSEThere are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSEOne Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSEExplain zero factorial class 11 maths CBSEState the laws of reflection of light10 examples of friction in our daily lifeWho is known as the father of chemistry class 11 chemistry CBSE

• 1
• 2

Repeaters Course for NEET 2022 - 23

NEET Repeater 2023 - Aakrosh 1 Year Course