During a titration experiment, 0.05 moles of carbon (lV) oxide is liberated. What is the volume of gas liberated?

In a titration experiment, a certain amount of a substance is reacted with a known volume of a solution of known concentration until the reaction is complete. The volume of the solution required to complete the reaction is measured, and this information is used to calculate the amount of substance present in the original sample. In this particular titration experiment, 0.05 moles of carbon (IV) oxide (CO2) is liberated. The question is asking for the volume of gas liberated, which means we need to find the volume of CO2 gas that corresponds to 0.05 moles. To do this, we can use the ideal gas law, which relates the volume of a gas to its pressure, temperature, and number of moles. The ideal gas law is expressed as: PV = nRT where P is the pressure of the gas, V is its volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature of the gas. Assuming that the experiment was carried out at standard temperature and pressure (STP), which is defined as 0°C and 1 atm pressure, we can simplify the ideal gas law to: V = nRT/P At STP, the pressure (P) is 1 atm and the temperature (T) is 0°C or 273.15 K. The universal gas constant (R) is 0.08206 L atm K^-1 mol^-1. Substituting these values into the ideal gas law, we get: V = (0.05 mol) x (0.08206 L atm K^-1 mol^-1) x (273.15 K) / (1 atm) Simplifying the expression gives us: V = 1.12 L Therefore, the volume of gas liberated in this titration experiment is 1.12 dm^3 or 1.12 liters. Option D is the correct answer, which states "1.12 dm^3".