When calcium carbonate is heated strongly, carbon dioxide gas is released.

CaCO3(s) CaO(s) + CO2(g)
What volume of CO2(g), measured at STP, is produced if 14.1 g of CaCO3(s) is heated?

You need to find the arrow key and use it.

mols CaCO3 = grams/molar mass
Using the coefficients in the balanced equation, convert mols CaCO3 to mols CO2. Now convert mols CO2 to grams CO2. g = mols x molar mass

iuiu

To calculate the volume of CO2 gas produced when 14.1 g of CaCO3 is heated, we need to use the ideal gas law and the molar volume at standard temperature and pressure (STP).

1. Start by calculating the number of moles of CaCO3.
- The molar mass of CaCO3 is 100.1 g/mol (40.1 g/mol for Ca + 12.0 g/mol for C + 3 * 16.0 g/mol for O).
- Number of moles = mass / molar mass = 14.1 g / 100.1 g/mol ≈ 0.141 mol.

2. According to the balanced chemical equation, 1 mole of CaCO3 produces 1 mole of CO2 gas.
So, the number of moles of CO2 produced is also 0.141 mol.

3. Now we need to use the molar volume of a gas at STP, which is 22.4 L/mol.
- Based on the ideal gas law, the volume (V) is given by V = n * Vm, where n is the number of moles and Vm is the molar volume.
- Therefore, volume of CO2 gas produced = 0.141 mol * 22.4 L/mol = 3.16 L.

Therefore, when 14.1 g of CaCO3 is heated strongly, approximately 3.16 L of CO2 gas is produced at STP.

To find the volume of CO2 gas produced when 14.1 g of CaCO3 is heated, we need to follow these steps:

Step 1: Calculate the moles of CaCO3.
The molar mass of CaCO3 is 40.08 g/mol + 12.01 g/mol + 3(16.00 g/mol) = 100.09 g/mol.
To get moles, divide the mass of CaCO3 by its molar mass:
Moles of CaCO3 = 14.1 g / 100.09 g/mol = 0.1409 mol.

Step 2: Use the balanced equation to determine the ratio of moles of CaCO3 to moles of CO2.
From the balanced equation, we see that 1 mole of CaCO3 produces 1 mole of CO2.
Therefore, the moles of CO2 produced will also be 0.1409 mol.

Step 3: Calculate the volume of CO2 gas at STP.
At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters.
Therefore, the volume of CO2 gas produced will be:
Volume of CO2 gas = 0.1409 mol × 22.4 L/mol ≈ 3.15 L.

So, when 14.1 g of CaCO3 is heated, approximately 3.15 liters of CO2 gas is produced at STP.