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Chapter 12: Problem 62

Suppose a certain automobile engine has a cylinder with a volume of 500.0 \(\mathrm{mL}\) that is filled with air \((21 \%\) oxygen) at a tempera- ture of \(55^{\circ} \mathrm{C}\) and a pressure of 101.0 \(\mathrm{kPa}\) . What mass of octane must be injected to react with all of the oxygen in the cylinder? $$2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g)$$

Short Answer

Expert verified
The mass of octane which must be injected to react with all of the oxygen in the cylinder is 0.0368 g.

Step by step solution

01

Convert the temperature from Celsius to Kelvin

First, convert the given temperature from Celsius to Kelvin because gas laws require the temperature to be in Kelvin. The conversion formula is: \(T(K) = T(°C) + 273.15\). So, \(T(K) = 55°C + 273.15 = 328.15 K\).
02

Calculate the moles of Oxygen

The volume of the cylinder is given as 500.0 mL. But only 21% of it is oxygen, so the volume of oxygen = 0.21 * 500.0 mL = 105.0 mL. Using the ideal gas law (\(PV=nRT\)), we can find the number of moles of oxygen: \( n = PV / RT = (101.0 kPa * 105.0 ml) / (8.314 kPa L/mol K * 328.15 K) = 0.00403 mol \) of Oxygen.
03

Use the stoichiometry of the given reaction

By using the balanced chemical equation \(2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)\), we can determine the moles of octane that will react with all of the oxygen. From the balanced equation, 1 mol of octane reacts with 12.5 mol of oxygen. So, 0.00403 mol of oxygen require 0.00403 / 12.5 mol of octane.
04

Calculate the mass of Octane

The formula mass of octane (C8H18) is \(8 * 12.01 + 18 * 1.01 = 114.22 g/mol\). The required mass of octane = 0.00403 / 12.5 * 114.22 g/mol = 0.0368 g.

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