How many moles of cao form when 98.60 g caco3 decompose
CaCO3 = CaO + CO2
We are given the amount of CaCO3 to decompose when heated. This will be our starting point.
98.60 g CaCO3 ( 1 mol CaCO3 / 100.09 grams CaCO3) (1 mol CaO / 1 mol CaCO3 ) ( 56.08 g O2 / 1 mol CaO) = 55.25 g CaO
0.9851mol. This is the rt answer just did it
CaCO3 ---> CaO + CO2
2) Molar ratios
1 mol CaCO3 : 1 mol CaO : 1 mol CO2
3) Number of moles of reactant: 98.60 g CaCO3
3.1) Calculate the molecular mass of CaCO3:
40.08 g/mol + 12.00 g/mol + 3*16.00 g/mol = 100.08 g / mol
3.2) Calculate thte number of moles using the formula numbe of moles = mass in grams / molecular mass
=> number of moles = 98.60 g / 100.08 g / mol = 1.015 mol of CaCO3
4) Using the theoretical molar ratios you know 1 mole of CaCO3 produce 1 mol of each product, then with 1.015 mol of CaCO3 you will obtain 1.015 mol of CO, which rounded to 3 significant figures is 1.02 mol.
0.9852 moles of CaO
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.
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 .
no.of moles=mass /relative molecular mass