consider the five balanced chemical reactions listed below, all using o₂ as a reactant. normally, o₂ is an…

consider the five balanced chemical reactions listed below, all using o₂ as a reactant. normally, o₂ is an excess reagent for reactions because there is a large amount of it available in the air. for this question, suppose that you have 5.91 moles of o₂ and 1.76 moles of any of the other reactants in every equation. select those equations below in which o₂ would be the limiting reactant. choose one or more: a. h₂s(g)+2o₂(g)→so₃(g)+h₂o(g) b. 4nh₃(g)+5o₂(g)→4no(g)+6h₂o(g) c. c₃h₈(g)+5o₂(g)→3co₂(g)+4h₂o(g) d. c₆h₁₂o₆(s)+6o₂(g)→6co₂(g)+6h₂o(g) e. none because in all the reactions o₂ is in excess

consider the five balanced chemical reactions listed below, all using o₂ as a reactant. normally, o₂ is an excess reagent for reactions because there is a large amount of it available in the air. for this question, suppose that you have 5.91 moles of o₂ and 1.76 moles of any of the other reactants in every equation. select those equations below in which o₂ would be the limiting reactant. choose one or more: a. h₂s(g)+2o₂(g)→so₃(g)+h₂o(g) b. 4nh₃(g)+5o₂(g)→4no(g)+6h₂o(g) c. c₃h₈(g)+5o₂(g)→3co₂(g)+4h₂o(g) d. c₆h₁₂o₆(s)+6o₂(g)→6co₂(g)+6h₂o(g) e. none because in all the reactions o₂ is in excess

Answer

Explanation:

Step1: Define limiting reactant concept

The limiting reactant is the reactant that gets consumed first in a chemical reaction. We'll use mole - ratios to determine if $\ce{O2}$ is the limiting reactant.

Step2: Analyze reaction A

For $\ce{H2S(g) + 2O2(g)\rightarrow SO3(g)+H2O(g)}$, the mole - ratio of $\ce{H2S}$ to $\ce{O2}$ is $1:2$. Given 1.76 moles of $\ce{H2S}$, the moles of $\ce{O2}$ required is $2\times1.76 = 3.52$ moles. Since we have 5.91 moles of $\ce{O2}$, $\ce{O2}$ is in excess.

Step3: Analyze reaction B

For $\ce{4NH3(g)+5O2(g)\rightarrow4NO(g) + 6H2O(g)}$, the mole - ratio of $\ce{NH3}$ to $\ce{O2}$ is $4:5$. Given 1.76 moles of $\ce{NH3}$, the moles of $\ce{O2}$ required is $\frac{5}{4}\times1.76= 2.2$ moles. Since we have 5.91 moles of $\ce{O2}$, $\ce{O2}$ is in excess.

Step4: Analyze reaction C

For $\ce{C3H8(g)+5O2(g)\rightarrow3CO2(g)+4H2O(g)}$, the mole - ratio of $\ce{C3H8}$ to $\ce{O2}$ is $1:5$. Given 1.76 moles of $\ce{C3H8}$, the moles of $\ce{O2}$ required is $5\times1.76 = 8.8$ moles. Since we have 5.91 moles of $\ce{O2}$, $\ce{O2}$ is the limiting reactant.

Step5: Analyze reaction D

For $\ce{C6H12O6(s)+6O2(g)\rightarrow6CO2(g)+6H2O(g)}$, the mole - ratio of $\ce{C6H12O6}$ to $\ce{O2}$ is $1:6$. Given 1.76 moles of $\ce{C6H12O6}$, the moles of $\ce{O2}$ required is $6\times1.76=10.56$ moles. Since we have 5.91 moles of $\ce{O2}$, $\ce{O2}$ is the limiting reactant.

Answer:

C. $\ce{C3H8(g)+5O2(g)\rightarrow3CO2(g)+4H2O(g)}$, D. $\ce{C6H12O6(s)+6O2(g)\rightarrow6CO2(g)+6H2O(g)}$