What volume of carbon dioxide is produced when 36.7 g of calcium carbonate reacts completely according to the following reaction at 25oC and 1 atm?

Here is a worked example. Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To determine the volume of carbon dioxide produced when 36.7 g of calcium carbonate reacts completely, we first need to calculate the number of moles of calcium carbonate using its formula weight and the given mass.

The formula weight of calcium carbonate (CaCO3) can be calculated by summing the atomic masses of its constituent elements:

- 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.

Formula weight of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol)
= 100.09 g/mol

Next, we can calculate the number of moles of calcium carbonate:

Moles of CaCO3 = Mass of CaCO3 / Formula weight of CaCO3
= 36.7 g / 100.09 g/mol
= 0.3667 mol (rounded to four decimal places)

Based on the balanced chemical equation for the reaction, we know that 1 mole of calcium carbonate produces 1 mole of carbon dioxide. Therefore, the number of moles of carbon dioxide produced will also be 0.3667 mol.

Finally, to calculate the volume of carbon dioxide at 25°C and 1 atm, we need to use the ideal gas equation:

PV = nRT

Where:
P = pressure (1 atm)
V = volume (to be determined)
n = number of moles (0.3667 mol)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (25°C + 273.15 = 298.15 K)

Rearranging the ideal gas equation to solve for V:

V = (nRT) / P
= (0.3667 mol * 0.0821 L·atm/mol·K * 298.15 K) / 1 atm
= 9.075 L (rounded to three decimal places)

Therefore, the volume of carbon dioxide produced is approximately 9.075 liters at 25°C and 1 atm.