CaCO3---> CaO+CO2 How many moles of CaO form when 98.60g CaCO3 decompose? 98.60g CaCO3= ______ mol CaO

Answer:

0.9852 moles of CaO

Explanation:

Reaction equation for the decomposition of CaCO₃:

CaCO₃ → CaO + CO₂

The question asks how many moles of CaO form when 98.60g of CaCO₃ decompose.

We can see from the reaction equation that for every mol of CaCO₃, one mol of CaO will be produced (molar ratio 1:1)

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So first we need to calculate how many moles are the 98.60g of CaCO₃:

Molar Mass of CaCO₃ = molar mass Ca + molar mass C + 3 * molar mass O

                                     = 40.078 + 12.011 + 3 * 15.999 = 100.086 g/mol

Moles of CaCO₃ = mass CaCO₃ / molar mass CaCO₃

Moles of CaCO₃ = 98.60 g / 100.086 g/mol = 0.9852 moles CaCO₃

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As we said before for every mol of CaCO₃, one mol of CaO is produced.

So the decomposition of 0.9852 moles of CaCO₃ will produce 0.9852 moles of CaO.

Hey I think it is going to be 0986 moles. Cuz we can see that no. of moles of CaCO3 which will decompose is equivalent to the no. of CaO . Now it's just the matter of finding the no. of moles of CaCO3 .