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

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

How many grams of calcium carbonate are needed to produce 59.0L of carbon dioxide at STP?

mb.b.

1. Write the balanced equation.

2. Convert 59.0 L CO2 at STP to moles. moles = 59.0 L x (1 mole/22.4 L) = ??
3. Using the coefficients in the balanced equation, convert moles CO2 to moles CaCO3.
4. Convert moles CaCO3 to grams.
grams = moles x molar mass.

To determine the grams of calcium carbonate needed to produce 59.0L of carbon dioxide at STP (Standard Temperature and Pressure), we need to use the stoichiometry of the reaction.

First, let's find the number of moles of carbon dioxide produced:
Using the ideal gas law, we know that at STP (0°C and 1 atm pressure), 1 mole of any gas occupies 22.4L of volume.
So, the number of moles of CO2 can be calculated as follows:

Number of moles of CO2 = (Volume of CO2 / Volume of 1 mole of CO2 at STP)
= (59.0 L / 22.4 L/mol)
= 2.63 moles of CO2

From the balanced equation, we can see that the ratio of moles of CO2 to moles of CaCO3 is 1:1. Therefore, the number of moles of CaCO3 required is also 2.63 moles.

Next, we need to convert moles of CaCO3 to grams using the molar mass (atomic weights) of the elements involved.
The molar mass of CaCO3 is:
1 mole of Ca = 40.08 g/mol
1 mole of C = 12.01 g/mol
3 moles of O = 16.00 g/mol x 3 = 48.00 g/mol
Total molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + 48.00 g/mol = 100.09 g/mol

Now, we can calculate the grams of CaCO3 needed:

Grams of CaCO3 = (Number of moles of CaCO3) x (Molar mass of CaCO3)
= (2.63 moles) x (100.09 g/mol)
= 263.27 g

Therefore, approximately 263.27 grams of calcium carbonate are needed to produce 59.0L of carbon dioxide at STP.

To determine the number of grams of calcium carbonate needed to produce 59.0L of carbon dioxide at STP (Standard Temperature and Pressure), we need to use the ideal gas law and stoichiometry.

Step 1: Convert the volume of carbon dioxide to moles.

To do this, we need to know the molar volume of a gas at STP, which is 22.4L. We can use the following conversion factor:

1 mole of any gas = 22.4L

So, the number of moles of carbon dioxide is given by:

Number of moles of CO2 = Volume of CO2 (L) / 22.4L

Number of moles of CO2 = 59.0L / 22.4L = 2.63 moles of CO2

Step 2: Use stoichiometry to convert moles of carbon dioxide to moles of calcium carbonate.

From the balanced chemical equation, we know that the stoichiometric ratio between calcium carbonate (CaCO3) and carbon dioxide (CO2) is 1:1. This means that 1 mole of calcium carbonate produces 1 mole of carbon dioxide. Therefore, the number of moles of calcium carbonate is also 2.63 moles.

Step 3: Convert moles of calcium carbonate to grams.

To convert moles to grams, we need to know the molar mass of calcium carbonate. The molar mass of calcium carbonate (CaCO3) is calculated by adding the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms:

Molar mass of CaCO3 = (atomic mass of Ca) + (atomic mass of C) + (3 x atomic mass of O)

Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol)

Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + 48.00 g/mol

Molar mass of CaCO3 = 100.09 g/mol

Now, we can calculate the mass of calcium carbonate:

Mass of CaCO3 = Number of moles of CaCO3 x Molar mass of CaCO3

Mass of CaCO3 = 2.63 moles x 100.09 g/mol

Mass of CaCO3 = 263.27 grams

Therefore, 263.27 grams of calcium carbonate are needed to produce 59.0L of carbon dioxide at STP.