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goldenfox [79]
4 years ago
15

A major component of gasoline is octane C8H18. When liquid octane is burned in air it reacts with oxygen O2 gas to produce carbo

n dioxide gas and water vapor. Calculate the moles of carbon dioxide produced by the reaction of 0.085mol of oxygen. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
Chemistry
1 answer:
Nady [450]4 years ago
7 0

Answer:

0.054 moles of carbon dioxide will be produced

Explanation:

Step 1: Data given

Number of moles oxygen = 0.085 moles

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles of carbon dioxide

For 2 moles octane we need 25 moles oxygen gas to produce 16 moles carbon dioxide gas and 18 moles water vapor

0.0544

For 0.085 moles O2 we'll have 16/25 * 0.085 = 0.054 moles CO2

0.054 moles of carbon dioxide will be produced

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<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.

<u>Explanation:</u>

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To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}

where,

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t = time taken = 1.52 hrs

[A_o] = Initial concentration of reactant = 100 g

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Putting values in above equation, we get:

k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}

To calculate the half life period of first order reaction, we use the equation:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life period of first order reaction = ?

k = rate constant = 0.2098hr^{-1}

Putting values in above equation, we get:

t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs

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Answer:

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