dummythicc2006
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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

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