2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g)If you burned one gallon of gas (C8H18) (approximately 4000 grams), how many liters of carbon dioxide would be produced at a temperature of 89.7°C and a pressure of 5.78 atm?
Question
2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g)If you burned one gallon of gas (C8H18) (approximately 4000 grams), how many liters of carbon dioxide would be produced at a temperature of 89.7°C and a pressure of 5.78 atm?
Solution
To solve this problem, we need to use the ideal gas law and stoichiometry.
Step 1: Convert grams of gasoline to moles. The molar mass of C8H18 is approximately 114 g/mol. So, 4000 g of C8H18 is 4000 g / 114 g/mol = 35.09 moles of C8H18.
Step 2: Use the balanced chemical equation to find moles of CO2 produced. According to the balanced chemical equation, 2 moles of C8H18 produce 16 moles of CO2. So, 35.09 moles of C8H18 will produce 35.09 moles * (16 moles CO2 / 2 moles C8H18) = 280.7 moles of CO2.
Step 3: Use the ideal gas law to find the volume of CO2 at the given temperature and pressure. The ideal gas law is PV = nRT, where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin. First, convert the temperature from Celsius to Kelvin. T = 89.7°C + 273.15 = 362.85 K. The ideal gas constant R is 0.0821 L·atm/(K·mol) when pressure is in atm and volume is in liters. So, V = nRT / P = 280.7 moles * 0.0821 L·atm/(K·mol) * 362.85 K / 5.78 atm = 1740 liters of CO2.
So, burning one gallon of gasoline would produce approximately 1740 liters of carbon dioxide at a temperature of 89.7°C and a pressure of 5.78 atm.
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