A sample of 30.695 g of limestone (calcium carbonate) is heated at high temperature and decomposes according to the calcination reaction

After a warm-up, the limestone suffered complete decomposition and after the cold was weighed and indicated 30.14 g . Determine the volume of gas released at STP

complete decomposition? I doubt it.

CaCO3>>CaO + CO2
so how many moles is (30.695-30.14)g?

moles=mass/44
lastly, volume at stp= moles*22.4 liters

I am not certain of the equation from your post. The warm up of what? The cold what was weighted? I will assume the reaction is

CaCO3(s) + heat ==> CaO + CO2
mass lost on heating is 30.695-30.14 = ? = mass CO2 produced.
mols CO2 = grams/molar mass = ? = n
You know 1 mol of gas @ STP has a volume of 22.4 L. Convert mols to L OR you may use the gas law, PV = nRT to solve for volume in liters.

To determine the volume of gas released at STP (Standard Temperature and Pressure), we need to calculate the number of moles of the gas first.

1. Calculate the molar mass of calcium carbonate (CaCO3):
Calcium (Ca) has a molar mass of 40.08 g/mol.
Carbon (C) has a molar mass of 12.01 g/mol.
Oxygen (O) has a molar mass of 16.00 g/mol.
So, the molar mass of calcium carbonate (CaCO3) is:
(1 x 40.08 g/mol) + (1 x 12.01 g/mol) + (3 x 16.00 g/mol) = 100.09 g/mol.

2. Calculate the number of moles of calcium carbonate decomposed:
Moles = Mass / Molar mass.
Moles = 30.695 g / 100.09 g/mol = 0.3067 mol (rounded to 4 decimal places).

3. According to the balanced chemical equation for the decomposition of calcium carbonate:
CaCO3 (s) → CaO (s) + CO2 (g).
One mole of calcium carbonate produces one mole of carbon dioxide gas.

4. Therefore, the number of moles of carbon dioxide gas released is also 0.3067 mol.

5. Finally, we can use the Ideal Gas Law equation to convert moles of gas to volume at STP:
PV = nRT,
where P is the pressure (at STP, it is 1 atm), 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 (273.15 K at STP).

6. Rearranging the equation to solve for V:
V = (nRT) / P,
V = (0.3067 mol) x (0.0821 L·atm/(mol·K)) x (273.15 K) / (1 atm).

7. Calculate the volume:
V = 7.15 L (rounded to 2 decimal places).

Therefore, the volume of gas released at STP is approximately 7.15 liters.