When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction

\rm CaCO_3{\it(s)}\rightarrow CaO{\it(s)}+CO_2{\it(g)}
How many grams of calcium carbonate are needed to produce 53.0 L of carbon dioxide at STP?

1. CaCO3(s) ==> CaO(s) + CO2(g)

2. Convert 53.0 L to moles. moles = 53.0/22.4.
3. Using the coefficients in the balanced equation, convert moles CO2 to moles CaCO3.
4. Now convert moles CaCO3 to grams. g = moles x molar mass.

To solve this problem, we need to calculate the number of moles of carbon dioxide gas produced and then convert it to grams of calcium carbonate.

Step 1: Convert the volume of carbon dioxide gas to moles.
According to the ideal gas law, at STP (Standard Temperature and Pressure), one mole of any gas occupies 22.4 liters. Therefore, we can calculate the number of moles of CO2 using the following equation:

n = V / Vm

Where:
n = number of moles
V = volume of CO2 (in liters)
Vm = molar volume of a gas at STP (22.4 L/mol)

Plugging in the values:
V = 53.0 L
Vm = 22.4 L/mol

n = 53.0 L / 22.4 L/mol
n = 2.37 mol

Step 2: Use the balanced equation to find the mole ratio between calcium carbonate and carbon dioxide.
By inspecting the balanced equation, we can see that for every 1 mole of CaCO3, 1 mole of CO2 is produced.

CaCO3(s) -> CaO(s) + CO2(g)

Step 3: Convert moles of CO2 to moles of CaCO3.
Since the mole ratio is 1:1 between CaCO3 and CO2, the number of moles of CaCO3 is the same as the number of moles of CO2.

Therefore, the number of moles of CaCO3 required is 2.37 mol.

Step 4: Convert moles of CaCO3 to grams.
To convert moles to grams, we need to use the molar mass of calcium carbonate (CaCO3). The molar mass of CaCO3 can be calculated by adding the molar masses of its individual elements.

CaCO3: Ca (40.08 g/mol) + C (12.01 g/mol) + 3O (16.00 g/mol) = 100.09 g/mol

Now, we can calculate the mass of calcium carbonate:

mass = n * molar mass
mass = 2.37 mol * 100.09 g/mol
mass = 237.23 g

Therefore, approximately 237.23 grams of calcium carbonate are needed to produce 53.0 L of carbon dioxide at STP.

To determine the number of grams of calcium carbonate needed to produce 53.0 L of carbon dioxide at STP (Standard Temperature and Pressure), we need to use the principles of stoichiometry.

The given balanced equation is:
CaCO₃(s) → CaO(s) + CO₂(g)

From the balanced equation, we can see that the ratio of calcium carbonate to carbon dioxide is 1:1. This means that for every 1 mole of calcium carbonate, we will produce 1 mole of carbon dioxide.

At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 L of volume.

Therefore, to find the number of moles of carbon dioxide gas produced, we can use the following formula:
Moles of gas = Volume of gas (in liters) / 22.4 L/mol

Given that the volume of carbon dioxide gas is 53.0 L, we can calculate the number of moles as follows:
Moles of CO₂ = 53.0 L / 22.4 L/mol ≈ 2.37 mol

Since the stoichiometric ratio between calcium carbonate and carbon dioxide is 1:1, we know that the number of moles of calcium carbonate required is also 2.37 mol.

Now, to find the mass of calcium carbonate, we need to use its molar mass. The molar mass of calcium carbonate (CaCO₃) can be calculated by adding up the atomic masses of its constituent elements: calcium (Ca), carbon (C), and oxygen (O).

The atomic mass of calcium (Ca) is 40.08 g/mol.
The atomic mass of carbon (C) is 12.01 g/mol.
The atomic mass of oxygen (O) is 16.00 g/mol.

Adding them up:
Molar mass of CaCO₃ = (40.08 g/mol) + (12.01 g/mol) + 3 × (16.00 g/mol) = 40.08 g/mol + 12.01 g/mol + 48.00 g/mol = 100.09 g/mol

Finally, we can calculate the mass of calcium carbonate needed using the following formula:
Mass = Moles × Molar mass

Mass = 2.37 mol × 100.09 g/mol ≈ 237 g

Therefore, approximately 237 grams of calcium carbonate are needed to produce 53.0 L of carbon dioxide at STP.

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