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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 61.0L of carbon dioxide at STP?
Express your answer with the appropriate units.

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Answer:

272.2 grams is the mass of calcium carbonate needed to produce 61.0L of carbon dioxide at STP

Explanation:

If the gas is produced at STP, by the volume and the Ideal Gas Law, we can know the moles.

P . V = n. R .T

1 atm . 61 L = n . 0.082L.atm/mol.K . 273K

61 L.atm / (0.082 mol.K/L.atm . 273K) = 2.72 moles

As the relation is 1:1

1 mol of CO₂ cames from 1 mol of CaCO₃, so 2.72 moles of CO₂ comes from 2.72 moles of CaCO₃

Molar mass CaCO₃ = 100.08 g/m

Moles . molar mass = grams

2.72 m . 100.08 g/m = 272.7 g

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