How many moles of cao form when 98.60 g caco3 decompose

How many moles of cao form when 98.60 g caco3 decompose

Answers

The balanced reaction is:

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

1) Chemical equation

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.

1.02 mol

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)

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

Its 0.9851 I just took it 

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