2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g)The above reaction is the reaction between gasoline (octane) and oxygen that occurs inside automobile engines. If 88.51 g of gasoline are burned, what volume, in L, of oxygen is needed if the pressure is 6.8 atm, and the temperature is 83.3°C?
Question
2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g)The above reaction is the reaction between gasoline (octane) and oxygen that occurs inside automobile engines. If 88.51 g of gasoline are burned, what volume, in L, of oxygen is needed if the pressure is 6.8 atm, and the temperature is 83.3°C?
Solution
To solve this problem, we need to use the ideal gas law equation, which is PV = nRT. But first, we need to find the number of moles of gasoline (octane) and oxygen.
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Calculate the molar mass of C8H18 (gasoline): C: 12.01 g/mol x 8 = 96.08 g/mol H: 1.01 g/mol x 18 = 18.18 g/mol Molar mass of C8H18 = 96.08 g/mol + 18.18 g/mol = 114.26 g/mol
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Calculate the number of moles of C8H18: n = mass/molar mass = 88.51 g / 114.26 g/mol = 0.774 moles of C8H18
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From the balanced chemical equation, we know that 2 moles of C8H18 react with 25 moles of O2. So, we can set up a ratio to find the number of moles of O2 needed: (25 moles O2 / 2 moles C8H18) x 0.774 moles C8H18 = 9.675 moles of O2
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Now we can use the ideal gas law to find the volume of O2. We need to convert the temperature to Kelvin first: T = 83.3°C + 273.15 = 356.45 K
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Substitute the known values into the ideal gas law: V = nRT/P = (9.675 moles * 0.0821 Latm/molK * 356.45 K) / 6.8 atm = 439.6 L
So, 439.6 L of oxygen is needed to burn 88.51 g of gasoline under the given conditions.
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