dummythicc2006
Answered

Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

PLEASE HELP FAST! When 23.3 g of O2 reacts with 18.3 g C10H8, what is the limiting reactant. The equation is C10H8 + 12O2 --> 10CO2 + 4H2O?

Choices:
A. C10H8
B. O2
C. CO2
D. H2O

Sagot :

Answer:

B

Explanation:

The answer is B, O2

Tutorconsortium330

O₂ is the limiting reagent.

What is a limiting reactant?

A limiting reactant is described as the one that will be consumed first in a chemical reaction.

Calculation

C₁₀H₈ + 12O₂ → 10CO₂ + 4H₂O

Given, Mass of O₂ = 23.3 g

           Mass of C₁₀H₈ = 18.3 g

Molar mass of O₂ = 32 g

No. of moles of O₂ = 32/23.3 = 1.4 moles

Molar mass of C₁₀H₈ = 128 g

No. of moles of C₁₀H₈ = 128/18.3 = 6.9 = 7 moles

According to the equation,

1 mole of C₁₀H₈ yields 10 moles of CO₂. So, 7 moles of C₁₀H₈, give 70 moles of CO₂.

12 moles of O₂ yields 10 moles of CO₂. So, 1.4 moles of O₂ give 1.16 moles of CO₂.

We do know, however, that based on our balanced equation, if we were to totally react all 1.4 moles of O2, we could only produce a maximum of 1.16 moles of CO2. Even while there is enough C10H8 to generate more, the amount of O2 we have is a limiting factor because it will be consumed first.

So, our limiting reactant is O₂.

Learn more about limiting reagents here:

https://brainly.com/question/11848702

#SPJ2

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.