Question

What mass of calcium oxide is obtained from thermal decomposition of 50g of calcium carbonate, assuming a 40% yield

Answer 1

Answer:

CaCO3 decomposition:

CaCO3 -> CaO + CO2

The first thing to do here is to calculate the relative formula mass of CaCO3 and CO2.

RFM of CaCO3 = 40 g + 12 g + (16 g x 3) = 100 g

RFM of CO2 = 12 g + (16 g x 2) = 44 g

The decomposition equation shows 1 CaCO3 goes to 1 CaO and 1 CO2.

We have 50 g of CaCO3, effectively we have half a CaCO3.

50 g / 100 g = 0.5

In order to get the mass of CO2 we need to times the RFM of CO2 by 0.5 (or divide by 2) because of the ratio of CaCO3 to CO2:

44 g x 0.5 = 22 g

22 g of CO2 are obtained from decomposition of 50 g of CaCO3.

Answer 2

Answer:

28 grams if you use whole numbers.

Explanation:

First, write reaction equation: CaCO3 ------> CaO + CO2

It is already balanced.

Second, you have to calculate how many moles of CaCO3 do you have.

It's easy: n(CaCO3)=m(CaCO3)/M(CaCO3) where M is the molar mass in g/mol

n= 50g / 100g/mol = 0.5 mol

Third, find how many moles of product do you get, in this case also 0.5 moles.

n(CaCO3)=n(CaO)

m(CaO) = n(CaO)*M(CaO) = 0.5mol*56g/mol = 28g.

Hope it helps you.