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dexar [7]
4 years ago
14

Here is the combustion reaction for octane (c8h18), which is a primary component of gasoline. how many moles of co2 are emitted

into the atmosphere when 15.1 g of c8h18 is burned?
Chemistry
1 answer:
Ivenika [448]4 years ago
3 0
<span>Combustion reaction for octane is as follows.
2C</span>₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Moles (mol) = mass (g) / molar mass (g/mol)

mass of octane = 15.1 g
molar mass of octane = <span>114.23 g/mol
Hence, moles of octane = 15. 1 g / </span><span>114.23 g/mol
                                       = 0.132 mol

The stoichiometric ratio between </span>C₈H₁₈ and CO₂ is 1 : 8.
Hence,
    moles of CO₂ produced = moles of octane burnt x 8
                                          = 0.132 mol x 8
                                          = 1.056 mol

Hence, 15.1 g of octane produce 1.056 moles of CO₂.
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Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

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The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

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Answer:

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Explanation:

<em>A chemist prepares a solution of potassium dichromate by measuring out 13.1 g of potassium dichromate into a 150 mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's potassium dichromate solution. Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Calculate the moles corresponding to 13.1 g of potassium dichromate

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13.1 g × (1 mol/294.19 g) = 0.0445 mol

Step 2: Convert the volume of solution to L

We will use the relationship 1 L = 1000 mL.

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Step 3: Calculate the concentration of the solution in mol/L

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