Question

Quick lime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO3) . Calculate the volume CO2 produced at STP from the decomposition of 152 g of CaCO3 according to the reaction.
CaCO3 (s) = CaO (s) + CO2(g)

Answer 1

First, we need to determine the number of moles of CaCO3 present in 152 g:

Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3(16.00 g/mol) = 100.09 g/mol

Number of moles = mass / molar mass = 152 g / 100.09 g/mol = 1.519 mol

According to the balanced equation, 1 mole of CaCO3 produces 1 mole of CO2.

Therefore, 1.519 moles of CaCO3 will produce 1.519 moles of CO2.

Now, we need to calculate the volume of CO2 at STP (Standard Temperature and Pressure) which is 0°C (273 K) and 1 atm pressure.

1 mole of any gas at STP occupies 22.4 L.

Therefore, 1.519 moles of CO2 will occupy:
Volume = 1.519 moles * 22.4 L/mole = 34.02 L

So, 152 g of CaCO3 will produce 34.02 L of CO2 at STP.