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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?

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Answer:

1.06 liters of gas are produced

Explanation:

Our reaction is:

CaCO₃(s) → CaO(s) + CO₂ (g)

This is the decomposition of calcium carbonate.

Ratio is 1:1:1

1 mol of carbonate can decompose to 1 mol of oxide and 1 mol of oxygen.

We convert mass to moles:

4.74 g . 1 mol /100.08g = 0.0474 moles

These are the moles of oxygen produced.

We know that 1 mol of any gas at STP is contained in 22.4L

0.0474 mol . 22.4L /mol = 1.06 L

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