How much energy in (kj) is required to decompose 14.7 g of caco3.

caco3(s) + 556 kj > cao(s) + co2(g)

81.732KJ

Explanation:

Reaction : CaCO₃(s) → CaO(s) + CO₂(g), +556KJ

⇒To decompose 1 mole of CaCO₃, 556KJ of energy is required

The molecular weight of CaCO₃ is 40 + 12 + 3×16

⇒ molecular weight of CaCO₃ = 100 g

∴ 100 g of CaCO₃ requires 556KJ of energy

Need to find out how much energy is required by 14.7 g of CaCO₃

⇒ =

⇒ E = ×14.7 KJ = 81.732 KJ

∴ Energy required to decompose 14.7 g of CaCO₃ is 81.732 KJ

81.732 kilo Joules of energy is required to decompose 14.7 grams of calcium carbonate.

Explanation:

Moles of calcium carbonate =

According to reaction, 1 mole of calcium carbonate requires 556 kilo Joules of energy to decompose:

Then 0.147 moles of calcium carbonate will need:

81.732 kilo Joules of energy is required to decompose 14.7 grams of calcium carbonate.

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