When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction CaCO3(s)→CaO(s)+CO2(g)

What is the mass of calcium carbonate needed to produce 55.0 L of carbon dioxide at STP?
Express your answer with the appropriate units.

mass of CaCO3 =

The mass of Calcium carbonate needed : 246 g

Reaction

CaCO₃(s)→CaO(s)+CO₂(g)

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

Then for 55 L of CO₂, mol :

From the equation, mol ratio for mol CaCO₃ : mol CO₂ = 1 : 1, so :

mol CaCO₃ = mol CO₂ = 2.46

mass CaCO₃(MW=100g/mol) :

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

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