What volume of carbon dioxide would evolve from fierce heating of a #200*g# mass of #"calcium carbonate"# on heating under standard conditions of #"STP"#?

Question

Jeremiah Adams · Accepted Answer

Well, #"STP"# specifies a temperature of #273.15*K# and an absolute pressure of exactly #100 *kPa#. I gets under #50*L#.

We investigate the breakdown reaction. #CaCO_3(s) + Delta rarrCaO(s) + CO_2(g)uarr# And given complete decomposition we have a molar quantity of #(200*g)/(100.09*g*mol^-1)=2.0*mol# WITH RESPECT TO CALCIUM CARBONATE, AND CARBON DIOXIDE. And as you know this molar quantity also represents a (large!) number with respect to gaseous molecules. And the volume expressed by this quantity is simply given by the old Ideal Gas equation..#V=(nRT)/P# #-=(2.0*molxx0.0821*(L*atm)/(K*mol)xx273.15*K)/((100*kPa)/(101.3*kPa*atm^-1))# #=??*L# And the number of molecules is simply #2*molxx6.022xx10^23*mol^-1=??#

Avery Adams · Answer

undefined To calculate the volume of carbon dioxide evolved from heating 200 g of calcium carbonate at STP, you use the stoichiometry of the reaction. The balanced chemical equation for the reaction is:

CaCO3(s) → CaO(s) + CO2(g)

The molar mass of calcium carbonate (CaCO3) is 100.09 g/mol, and the molar mass of carbon dioxide (CO2) is 44.01 g/mol.

Using the stoichiometry of the reaction, you find that 1 mole of calcium carbonate produces 1 mole of carbon dioxide.

First, convert the mass of calcium carbonate to moles:
200 g / 100.09 g/mol = 1.999 moles

Since the mole ratio of calcium carbonate to carbon dioxide is 1:1, the number of moles of carbon dioxide produced is also 1.999 moles.

Now, use the ideal gas law to find the volume of carbon dioxide:
PV = nRT

At STP, P = 1 atm, T = 273 K, and R = 0.0821 L·atm/mol·K.

V = (1.999 moles) * (0.0821 L·atm/mol·K) * (273 K / 1 atm) = 45.33 L

Therefore, approximately 45.33 liters of carbon dioxide would evolve from heating 200 g of calcium carbonate at STP.