The volume of co2 gas produced at NTP when 25 gram of caco3 is heated
22.4 liters/mol
calcium carbonate is 100 g/mol
so we have 1/4 mol
(1/4)(22.4) = 5.6 liters = .0056 m^3
To find the volume of CO2 gas produced when 25 grams of CaCO3 is heated at NTP (Normal Temperature and Pressure), you need to use a few steps:
1. Calculate the moles of CaCO3: Divide the given mass of CaCO3 (25 grams) by its molar mass. The molar mass of CaCO3 is calculated by adding the atomic masses of its constituent elements (Ca: 40.08 g/mol, C: 12.01 g/mol, O: 16.00 g/mol). Thus, the molar mass of CaCO3 is 40.08+12.01+(16.00*3) = 100.09 g/mol. Therefore, the number of moles of CaCO3 is 25 g / 100.09 g/mol = 0.2499 moles (approximately).
2. Determine the stoichiometric ratio: Balanced chemical equation shows that when CaCO3 decomposes, 1 mole of CaCO3 produces 1 mole of CO2. Therefore, the number of moles of CO2 gas produced is also 0.2499 moles.
3. Apply the Ideal Gas Law to calculate the volume of CO2 gas: The Ideal Gas Law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L*atm/mol*K), and T is the temperature in Kelvin.
At NTP, the pressure (P) is 1 atm, and the temperature (T) is 273.15 K.
Substituting the given values into the formula:
V = (nRT) / P
= (0.2499 moles * 0.0821 L*atm/mol*K * 273.15 K) / 1 atm
= 5.94 L (approximately)
Therefore, when 25 grams of CaCO3 is heated at NTP, it produces approximately 5.94 liters of CO2 gas.