when calcium carbonate, caco3(s) is heated, it decomposes to form calcium oxide, cao(s) and carbon dioxide gas. how many liters of carbon dioxide will be produced at stp if 2.38 kg of calcium carbonate reacts completely?

To determine the number of liters of carbon dioxide that will be produced at STP (Standard Temperature and Pressure), we need to follow these steps:

Step 1: Write and balance the chemical equation for the reaction:
CaCO3(s) -> CaO(s) + CO2(g)

Step 2: Calculate the moles of CaCO3:
Given that the mass of CaCO3 is 2.38 kg, we convert it to grams:
2.38 kg * 1000 g/kg = 2380 g

To find the moles of CaCO3, we need to divide the mass by the molar mass of CaCO3:
Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3(16.00 g/mol) (O) = 100.09 g/mol

Moles of CaCO3 = 2380 g / 100.09 g/mol

Step 3: Use the mole ratio to find the moles of CO2 produced:
From the balanced equation, we can see that for every 1 mole of CaCO3, 1 mole of CO2 is produced.

Moles of CO2 = Moles of CaCO3

Step 4: Use the ideal gas law to calculate the volume of CO2 at STP:
The ideal gas law equation is:
PV = nRT

At STP, the pressure (P) is 1 atm and temperature (T) is 273 K.

R is the ideal gas constant, which is 0.0821 L·atm/(mol·K).

V is the volume of the gas.

We use the rearranged form of the ideal gas law equation:

V = (nRT) / P

Convert the number of moles of CO2 to volume using the ideal gas law equation:

V (CO2) = (Moles of CO2 * R * T) / P

Step 5: Plug in the values and solve the equation:
Moles of CO2 = 2380 g / 100.09 g/mol
T = 273 K
P = 1 atm
R = 0.0821 L·atm/(mol·K)

V (CO2) = (2380 g / 100.09 g/mol) * (0.0821 L·atm/(mol·K)) * 273 K / 1 atm

Calculating this equation will give us the volume of CO2 in liters at STP.

To determine the volume of carbon dioxide produced at STP (standard temperature and pressure), we need to use the ideal gas law, which states that PV = nRT, where P is the pressure, 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.

First, let's calculate the number of moles of calcium carbonate (CaCO3) using its molar mass:

Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + (3 * 16.00 g/mol) (O)
= 100.09 g/mol

Number of moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
= 2.38 kg * (1000 g/kg) / 100.09 g/mol
= 23.78 mol

According to the balanced chemical equation, 1 mole of CaCO3 produces 1 mole of CO2.

So, the number of moles of CO2 produced = number of moles of CaCO3 = 23.78 mol

Now, let's calculate the volume of CO2 at STP. Standard temperature is 0°C or 273.15 K, and the standard pressure is 1 atm.

Using the ideal gas law, we can rearrange it to solve for V: V = nRT / P

V = (23.78 mol) * (0.0821 L.atm/mol.K) * (273.15 K) / (1 atm)
V ≈ 579.54 L

Therefore, approximately 579.54 liters of carbon dioxide will be produced at STP when 2.38 kg of calcium carbonate reacts completely.

I know it's cool to use lower case for everything; however, CO, Co and co mean different things in chemistry. When you type chemistry questions you should find the caps key and use it.


CaCO3 ==> CaO + CO2
mols CaCO3 = grams/molar mass = ?
Use the coefficients in the balanced equation to convert mols CaCO3 to mols CO2.
Then convert mols CO2 to L knowing that 1 mol occupies 22.4L at STP.