Answer:
0.9852 moles of CaO
Step-by-step 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)
_____________________________________________________
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
________________________________________________________
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.