Calculate volume of Carbon dioxide produced on heating 10g of limestone at S.T.P
To calculate the volume of carbon dioxide produced when heating limestone, we need to use the balanced chemical equation and the ideal gas law.
First, we need to write the balanced chemical equation for the reaction of limestone (calcium carbonate) with heat:
CaCO3(s) → CaO(s) + CO2(g)
From the equation, we can see that 1 mol of limestone produces 1 mol of carbon dioxide gas. We need to find the number of moles of carbon dioxide produced from the given mass of limestone.
To do this, we need to consider the molar mass of CaCO3. The molar mass of carbon is 12.01 g/mol, the molar mass of oxygen is 16.00 g/mol, and the molar mass of calcium is 40.08 g/mol.
Molar mass of CaCO3 = (12.01 * 1) + (16.00 * 3) + (40.08 * 1) = 100.09 g/mol
Next, we can calculate the number of moles of CaCO3 using the given mass of 10 g:
Number of moles = Mass / Molar mass
Number of moles = 10 g / 100.09 g/mol = 0.0999 mol (approximately)
Since the reaction is at STP (Standard Temperature and Pressure), we can use the ideal gas law to calculate the volume of carbon dioxide:
PV = nRT
Where:
P = Pressure (at STP, it is 1 atm)
V = Volume (what we need to find)
n = Number of moles (0.0999 mol)
R = Ideal Gas Constant (0.0821 L atm / (mol K))
T = Temperature (at STP, it is 273 K)
Rearranging the equation, we have:
V = (n * R * T) / P
V = (0.0999 mol * 0.0821 L atm / (mol K) * 273 K) / 1 atm
V ≈ 2.17 L
Therefore, approximately 2.17 liters of carbon dioxide gas is produced by heating 10 grams of limestone at STP.
22.4 liters/mol
CO2 = 12 + 16 + 16 = 44 grams/mol
so
(10/44)22.4 = 5.1 liters