What is the mass of Caco3 containing 67.2 liters of o2 at STP?
67.2/22.4 = 3 moles of O2
So that will be 2 moles of CaCO3
how many grams is that?
The molar mass of CaCO3 is approximately 100.09 g/mol (40.08 g/mol for calcium + 12.01 g/mol for carbon + 3 x 16.00 g/mol for oxygen). Therefore, 2 moles of CaCO3 would have a mass of:
2 moles CaCO3 x 100.09 g/mol = 200.18 g
So the mass of CaCO3 containing 67.2 L of O2 at STP would be 200.18 grams.
To determine the mass of CaCO3 containing 67.2 liters of O2 at STP (Standard Temperature and Pressure), you would need to follow these steps:
1. Determine the balanced chemical equation for the decomposition of CaCO3:
CaCO3 → CaO + CO2
2. Use the stoichiometry of the balanced equation to determine the molar ratio between O2 and CaCO3. In this case, for every 1 mole of CaCO3, one-third of a mole of O2 is consumed:
1 mole CaCO3 → 1/3 mole O2
3. Convert the given volume of O2 at STP into moles. At STP, one mole of any gas occupies 22.4 liters of volume:
67.2 liters O2 x (1 mole O2 / 22.4 liters O2) = 3 moles O2
4. Use the molar ratio from step 2 to calculate the moles of CaCO3 consumed:
3 moles O2 x (1 mole CaCO3 / 1/3 mole O2) = 9 moles CaCO3
5. Determine the molar mass of CaCO3. The molar mass of CaCO3 is the sum of the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms:
Molar mass CaCO3 = (1 x atomic mass Ca) + (1 x atomic mass C) + (3 x atomic mass O)
6. Multiply the moles of CaCO3 by the molar mass of CaCO3 to find the mass of CaCO3 consumed:
Mass CaCO3 = moles CaCO3 x molar mass CaCO3
By following these steps, you can calculate the mass of CaCO3 containing 67.2 liters of O2 at STP.