H 2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g)How many grams of water would be produced if 3.43 liters of oxygen were burned at a temperature of 55.791°C and a pressure of 1.15 atm?

To solve this problem, we need to use the ideal gas law and stoichiometry.

Step 1: Use the ideal gas law to find the number of moles of oxygen. The ideal gas law is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.

First, convert the temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is K = °C + 273.15. So, 55.791°C = 55.791 + 273.15 = 328.941 K.

Next, we need to use the correct value for R. Since the pressure is given in atm and the volume in liters, we use R = 0.0821 L·atm/K·mol.

Now we can solve for n (number of moles of oxygen): 1.15 atm * 3.43 L = n * 0.0821 L·atm/K·mol * 328.941 K n = (1.15 atm * 3.43 L) / (0.0821 L·atm/K·mol * 328.941 K) n = 0.144 moles of O2

Step 2: Use stoichiometry to find the number of moles of water produced. From the balanced chemical equation, we know that 25 moles of O2 produce 18 moles of H2O. So, 0.144 moles of O2 would produce: (0.144 moles O2 * 18 moles H2O) / 25 moles O2 = 0.10368 moles H2O

Step 3: Convert moles of water to grams. The molar mass of water (H2O) is approximately 18.015 g/mol. So, 0.10368 moles H2O * 18.015 g/mol = 1.87 g H2O

So, if 3.43 liters of oxygen were burned at a temperature of 55.791°C and a pressure of 1.15 atm, approximately 1.87 grams of water would be produced.