The thermal decomposition of calcium carbonate produces 2 by-products, c02 and calcium oxide. calculate the volume of the co2 produced at STP from the decomposition of 231 grams of caco3.

i know the equation i need to use is V=nrt/p but am i supposed to convert the grams to moles and use that for the value of n or do i need to come up with a mole ratio from the equation? i'm kind of stuck. all i have so far is
caco3 ----> cao + co2

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YOu are given the grams of calcium carbonate, convert that to moles. write the balanced equation, as you did, so you know the moles (n) of CO2. (for each mole of calcium carbonate, you get one mole of carbon dioxide).

Then use PV=nRT

To calculate the volume of CO2 produced at STP (Standard Temperature and Pressure) from the decomposition of 231 grams of CaCO3, you need to use stoichiometry and convert grams to moles. Here's how you can proceed:

1. Start with the balanced chemical equation:
CaCO3 ----> CaO + CO2

2. Determine the molar mass of CaCO3 and CO2:
- Molar mass of CaCO3 = 40.08 g/mol (Ca = 40.08 g/mol, C = 12.01 g/mol, O = 16.00 g/mol)
- Molar mass of CO2 = 44.01 g/mol (C = 12.01 g/mol, O = 16.00 g/mol)

3. Convert the given mass of CaCO3 to moles:
moles of CaCO3 = mass (g) / molar mass (g/mol)
moles of CaCO3 = 231 g / 100.09 g/mol = 2.309 mol

4. Since the stoichiometry of the reaction indicates that the molar ratio between CaCO3 and CO2 is 1:1, the moles of CO2 produced will also be 2.309 mol.

5. To calculate the volume of CO2 at STP, you can use the ideal gas law equation you mentioned:
V = nRT/P

- n = moles of CO2 = 2.309 mol
- R = ideal gas constant = 0.0821 Lยทatm/(molยทK)
- T = temperature at STP = 273.15 K
- P = pressure at STP = 1 atm

V = (2.309 mol)(0.0821 Lยทatm/(molยทK))(273.15 K) / 1 atm
V โ‰ˆ 52.4 L

Therefore, the volume of CO2 produced at STP from the thermal decomposition of 231 grams of CaCO3 is approximately 52.4 liters.