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vesna_86 [32]
3 years ago
7

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

Chemistry
2 answers:
kramer3 years ago
7 0

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)

_____________________________________________________

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

________________________________________________________

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.

Sav [38]3 years ago
3 0

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 .

<span>no.of moles=<span><span>mass /</span><span>relative molecular mass</span></span></span>

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