Answer:
0.9852 moles of CaO
Explanation:
Reaction equation for the decomposition of CaCO₃:
CaCO₃ → CaO + CO₂
The question asks <em>how many moles of CaO form when 98.60g of CaCO₃ decompose.</em>
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<u> calculate how many moles are the 98.60g of CaCO₃</u>:
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 <em>for every mol of CaCO₃, one mol of CaO is produced.</em>
So the decomposition of 0.9852 moles of CaCO₃ will produce 0.9852 moles of CaO.