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Ugo [173]
3 years ago
7

2b. 12 grams of C8H18 react with oxygen to produce carbon dioxide and water. How many grams of CO2 are produced?

Chemistry
1 answer:
likoan [24]3 years ago
4 0

Answer:

37.1g are produced

Explanation:

The combustion of C₈H₁₈ is:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

<em>Where 1 mole of C₈H₁₈ produce 8 moles of CO₂</em>

<em />

To find the mass of CO₂ that is produced we need to convert the mass of C₈H₁₈ with molar mass. Then, with the chemical equation, we can find the moles of CO₂ and its mass, as follows:

<em>Moles C₈H₁₈ -Molar mass: 114.2g/mol-</em>

12g C₈H₁₈ * (1mol / 114.2g) = 0.105 moles of C₈H₁₈

<em>Moles CO₂:</em>

As 1 mole of C₈H₁₈ produce 8 moles of CO₂, 0.105 moles of C₈H₁₈ produce:

0.105 moles of C₈H₁₈ * (8moles CO₂ / 1mole C₈H₁₈) = 0.84 moles of CO₂

<em>Mass CO₂ -Molar mass: 44.01g/mol-:</em>

0.84 moles of CO₂ * (44.01g / mol) = 37.0g of CO₂ ≈

<h3>37.1g are produced</h3>
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Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)

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t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}

[A]=0.065 M

0.065 M is its molarity after a reaction time of 19.0 h.

b)

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=x

Final concentration of [Co(NH_3)5Br]^{2+} after t = [A]=(100\%-69\%) x=31\%x=0.31x

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}

t = 185,902.06 s = \frac{185,902.06 }{3600} hour = 51.64 hours ≈ 52 hours

In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

6 0
4 years ago
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