How much of 80 % pure CaCO₃ will be required to produce 44.8 L of CO₂ at NTP ?

To determine the amount of 80% pure calcium carbonate (CaCO3) needed to produce 44.8 liters of carbon dioxide (CO2) at Normal Temperature and Pressure (NTP), we can start by understanding the chemical reaction involved. When calcium carbonate decomposes, it produces carbon dioxide and calcium oxide:

The Reaction Breakdown

The balanced chemical equation for the decomposition of calcium carbonate is:

• CaCO3 (s) → CaO (s) + CO2 (g)

This indicates that 1 mole of calcium carbonate produces 1 mole of carbon dioxide. At NTP, 1 mole of any ideal gas occupies 22.4 liters. Therefore, if we want to find out how many moles of CO2 are needed to produce 44.8 liters, we can use the following relationship:

Calculating Moles of CO2

Using the formula:

• Moles of CO2 = Volume of CO2 (L) / Volume of 1 mole at NTP (L)

Substituting in the values:

• Moles of CO2 = 44.8 L / 22.4 L/mol = 2 moles

Finding Moles of CaCO3 Required

Since the balanced equation tells us that 1 mole of CaCO3 produces 1 mole of CO2, this means we need 2 moles of CaCO3 to produce 2 moles of CO2.

Calculating Mass of Pure CaCO3

Next, we need to calculate the mass of pure CaCO3 required. The molar mass of CaCO3 is approximately 100 g/mol. Therefore, for 2 moles:

• Mass of CaCO3 = Moles × Molar mass = 2 moles × 100 g/mol = 200 g

Considering Purity

Since the calcium carbonate you have is only 80% pure, we must adjust our calculation to account for this purity level. This means that only 80% of the mass is actually CaCO3. To find the total mass of the 80% pure CaCO3 needed, we can use the following formula:

• Total mass = Mass of pure CaCO3 / Purity fraction

Substituting in the values:

• Total mass = 200 g / 0.80 = 250 g

Final Answer

Hence, to produce 44.8 liters of CO2 at NTP, you will need 250 grams of 80% pure calcium carbonate.