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erica [24]
2 years ago
6

A major component of gasoline is octane (C8H18). When octane is burned in air, it chemically reacts with oxygen gas (O2) to prod

uce carbon dioxide (CO2) and water (H2O).
What mass of carbon dioxide is produced by the reaction of 9.88 g of oxygen gas?

Round your answer to 3 significant digits.
Chemistry
1 answer:
Molodets [167]2 years ago
7 0

Answers:

8.70 g

Step-by-step explanation:

We know we will need a balanced equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.  

M_r:                   32.00      44.01

           2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 18H₂O

m/g:                    9.88

(a) Calculate the <em>moles of O₂ </em>

n = 9.88 g O₂ ×1 mol O₂ /32.00 g O₂

n = 0.3088 mol O₂

(b) Calculate the <em>moles of CO₂</em>

The molar ratio is (16 mol CO₂/25 mol O₂)

n = 0.3088 mol O₂ × (16 mol CO₂/25 mol O₂)

n = 0.1976 mol CO₂

(c) Calculate the <em>mass of CO₂ </em>

Mass of CO₂ = 0.1976 mol CO₂ × (44.01 g CO₂/1 mol CO₂)

Mass of CO₂ = 8.70 g CO₂

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The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

rate=k[H_2]^2[NH_3]

At a certain concentration of H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

0.120=k[H_2]^2[NH_3] ....(1)

Expression for rate law for second observation:

R=k(\frac{[H_2]}{2})^2[NH_3] ....(2)

Dividing 2 by 1, we get:

\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}

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Answer:

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Explanation:

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Explanation:

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Its atomic number is 20. So it has 20 protons and 20 electrons.

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So, you have two more electrons (20 - 18 = 2) to distribute.

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