When calcium carbonate is heated strongly, carbon dioxide gas is evolved.
CaCO3(s) —> CaO(s) + CO2 (g)
If 4.74 g of calcium carbonate is heated, what volume of CO2 (g) would be produced when collected at STP
To find the volume of CO2 gas produced at STP (Standard Temperature and Pressure), we need to use the ideal gas law:
PV = nRT
Where:
P = pressure (STP is 1 atm)
V = volume of gas in liters
n = number of moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (STP is 273 K)
First, we need to find the number of moles of CO2 produced when heating 4.74 g of calcium carbonate:
1. Calculate the molar mass of CaCO3:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Molar mass of CaCO3 = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol
2. Convert the mass of CaCO3 in grams to moles:
4.74 g / 100.09 g/mol = 0.0474 mol
Since the balanced chemical equation shows a 1:1 ratio between CaCO3 and CO2, 0.0474 moles of CaCO3 will produce 0.0474 moles of CO2.
3. Now, substitute the values into the ideal gas law:
PV = nRT
V = nRT / P
V = (0.0474 mol)(0.0821 L.atm/mol.K)(273 K) / 1 atm
V = 1.08 L
Therefore, when 4.74 g of calcium carbonate is heated and reacts to produce CO2 gas collected at STP, approximately 1.08 liters of CO2 gas would be produced.