How many grams of calcium carbonate are required to prepare for 50.0 grams of calcium oxide

the question is based on the decomposition of calcium carbonate through heating;

CaCO3(s) --> CaO(s) + CO2(g)

50 grams of calcium oxide is 0.893moles. the theoretical mole is 1:1 between caCO3 and CaO, so 0.893moles of CaCO3 is needed.

mass of CaCO3 = moles of CaCO3 x Molar mass of CaCO3

hope that helps...

To find out how many grams of calcium carbonate are required to prepare 50.0 grams of calcium oxide, we need to understand the stoichiometry of the reaction between calcium oxide (CaO) and calcium carbonate (CaCO3).

The balanced chemical equation for the reaction is:
CaCO3 --> CaO + CO2

From the equation, we can see that 1 mole of calcium oxide (CaO) is formed from 1 mole of calcium carbonate (CaCO3). The molar mass of calcium oxide (CaO) is 56.08 g/mol, while the molar mass of calcium carbonate (CaCO3) is 100.09 g/mol.

Now, let's calculate the number of moles of calcium oxide present in 50.0 grams using the equation:
moles = mass / molar mass

moles of calcium oxide = 50.0 g / 56.08 g/mol
moles of calcium oxide ≈ 0.892 mol

Since 1 mole of calcium carbonate is required to produce 1 mole of calcium oxide, we can determine the number of moles of calcium carbonate needed. Therefore, the number of moles of calcium carbonate required is also approximately 0.892 mol.

Finally, let's calculate the mass of calcium carbonate using the equation:
mass = moles × molar mass

mass of calcium carbonate = 0.892 mol × 100.09 g/mol
mass of calcium carbonate ≈ 89.44 grams

Therefore, approximately 89.44 grams of calcium carbonate are required to prepare 50.0 grams of calcium oxide.

Have you considered writing the balanced equation?

CaCO3>>CaO + O2

for each mole of calcium oxide, you need that number of moles of calcium carbonate.

a. how many moles of calcium oxide it 50 grams?
b. for that number of moles of calcium carbonate, how many grams is that?

@ bobpursley..your equation is incorrect so i came up with this one...

Bonjo, you are correct, I have a wrapped right thumb, messing up my typing, the correct equation is CO2 as a product, not O2. the rest of my response is correct.