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Chapter 10: Problem 106

Assume that a single cylinder of an automobile engine has a volume of 524 \(\mathrm{cm}^{3} .\) (a) If the cylinder is full of air at \(74^{\circ} \mathrm{C}\) and 0.980 atm, how many moles of \(\mathrm{O}_{2}\) are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 . )(\mathbf{b})\) How many grams of \(\mathrm{C}_{8} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2},\) assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ?

Short Answer

Expert verified
(a) There are approximately 0.00406 moles of O₂ present in the cylinder. (b) About 0.0371 grams of C₈H₁₈ could be combusted by this quantity of O₂, assuming complete combustion with the formation of CO₂ and H₂O.

Step by step solution

01

Convert the given cylinder volume to liters

The cylinder volume is given in cm³. We will convert it to L (liters) to match the units used for the Ideal Gas Law. \(Volume \ (L) = \frac{524 cm^3}{1000 cm^3/L} = 0.524 L\)
02

Calculate the total moles of air in the cylinder using the Ideal Gas Law

We will use the Ideal Gas Law (PV = nRT) to find the total number of moles of air (n) in the cylinder. \(n = \frac{PV}{RT}\) P = 0.980 atm (already in appropriate units) V = 0.524 L (found in Step 1) T = 74°C + 273.15 = 347.15 K (converted to Kelvin) R = 0.0821 L atm/(mol K) (Ideal Gas constant) \(n = \frac{(0.980 atm)(0.524 L)}{(0.0821 L atm/mol K)(347.15 K)} ≈ 0.0194 mol\)
03

Calculate the moles of O₂ present in the cylinder

Now we'll multiply the total moles of air by the mole fraction of O₂ (0.2095) to find the moles of O₂ present. Moles of O₂ = (0.0194 mol)(0.2095) ≈ 0.00406 mol
04

Determine how many moles of octane can be combusted by using the moles of O₂ and the balanced combustion reaction

The balanced combustion reaction is: C₈H₁₈ + 12.5 O₂ -> 8 CO₂ + 9 H₂O 1 mole of C₈H₁₈ requires 12.5 moles of O₂ for complete combustion. We have 0.00406 moles of O₂ present in the cylinder, so: Moles of C₈H₁₈ = (moles of O₂)/(12.5) Moles of C₈H₁₈ = (0.00406 mol) / (12.5) ≈ 0.000325 mol
05

Convert moles of C₈H₁₈ to grams

Lastly, we will convert the moles of C₈H₁₈ to grams using its molar mass (114.22 g/mol): Grams of C₈H₁₈ = (0.000325 mol)(114.22 g/mol) ≈ 0.0371 g Therefore, 0.0371 grams of C₈H₁₈ could be combusted by this quantity of O₂, assuming complete combustion with the formation of CO₂ and H₂O.

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Most popular questions from this chapter

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