When calcium carbonate is heated strongly, carbon dioxide gas is evolved.

Question

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

Answer 1

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.