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I'm not smart and I don't understand how to solve this problem:
Octane, one of the major components of gasoline, burns in air according to this unbalanced equation:
C₈H₁₈ (l) + O₂ (g) ----> CO₂ (g) +H₂O
The formula balanced is: 2 C₈H₁₈ (l) + 50 O₂ (g) ----> 16 CO₂ (g) + 18 H₂O
a) What volume of O₂ at STP is needed to burn 702g (1.00L) of octane?
b) What volume of O₂ at 18°C and 0.975 atm is needed to burn 702g of octane?

User Dkantowitz
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2 Answers

3 votes

Answer:

C₈H₁₈

Step-by-step explanation:

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User Bluegenetic
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4.7k points
3 votes

Answer:

a. 137.93 dm3

b. 150.61 L

User Bharat Sinha
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