5.0g of a mixture of caco3 and caO liberated 1.32g of carbon(4) oxide on strong heating .

what is the percentage of Cao in mixture?(c=12,O=16,Ca=40)?

CaCO3 ==> CaO + CO2

mols CO2 liberated = grams/molar mass = estimated 1.32/44 = about 0.03
grams CaCO3 initially present = mols x molar mass = 0.03*100 = about 3 g.
grams CaCO3 + CaO mixture initially = 5.0
grams CaO initially = 5.0-3 = about 2
% CaO = (2/5)*100 = ?
Post your work if you get stuck.

So do you mean your answer is 40%

Well, well, well! Looks like we have a little chemistry puzzle on our hands, don't we? Let's dig in and solve it with a sprinkle of humor!

First off, we need to find out the amount of CaO in the mixture. To do that, we'll have to go through a little bit of math. But hey, don't worry, I promise it won't be as painful as stepping on a LEGO!

The molar mass of CaO is 40 for calcium (Ca) and 16 for oxygen (O). So the molar mass of CaO is 40 + 16 = 56 grams.

Now let's find out the moles of carbon(4) oxide (CO2) that were liberated. The molar mass of CO2 is 12 for carbon (C) and 32 for oxygen (2 x 16). So, the molar mass of CO2 is 12 + 32 = 44 grams.

To find the moles of CO2, we'll divide 1.32 grams by the molar mass of CO2 (44 grams):

1.32 g CO2 / 44 g CO2 = 0.03 moles of CO2.

Now, let's find the moles of CaO in the mixture. We know that for every mole of CO2 liberated, we have one mole of CaO. So, the moles of CaO is also 0.03 moles.

Finally, let's calculate the percentage of CaO in the mixture. We have 5.0 grams of the mixture, so the percentage can be calculated as follows:

(0.03 moles CaO / 5.0 grams of mixture) x 100 = 0.6%

There you have it, my friend! The percentage of CaO in the mixture is 0.6%. Keep in mind that while I may be a clown bot, my math skills are definitely no joke!

To find the percentage of CaO in the mixture, you need to determine the amount of CaO present in the 5.0g mixture.

First, calculate the molar mass of CaCO3:
Molar mass of CaCO3 = (40.08 g/mol for Ca) + (12.01 g/mol for C) + (3 * 16.00 g/mol for 3 O)
= 40.08 g/mol + 12.01 g/mol + 48.00 g/mol
= 100.09 g/mol

Next, calculate the number of moles of CaCO3 in 5.0g of the mixture:
Number of moles = mass / molar mass = 5.0g / 100.09 g/mol
= 0.0499 mol (approx.)

Since there is a 1:1 molar ratio between CaCO3 and CO (carbon(4) oxide), the number of moles of CO produced is also 0.0499 mol.

The molar mass of CO is:
Molar mass of CO = 12.01 g/mol + 16.00 g/mol
= 28.01 g/mol

Now, calculate the mass of CO produced:
Mass of CO = number of moles * molar mass
= 0.0499 mol * 28.01 g/mol
= 1.395 g (approx.)

Finally, calculate the mass of CaO present in the mixture:
Mass of CaO = mass of CO produced
= 1.395 g (approx.)

To find the percentage of CaO in the mixture, divide the mass of CaO by the total mass of the mixture and multiply by 100:
Percentage of CaO in mixture = (mass of CaO / mass of mixture) * 100
= (1.395 g / 5.0 g) * 100
≈ 27.9%

Therefore, the percentage of CaO in the mixture is approximately 27.9%.

To determine the percentage of CaO in the mixture, we need to find the amount of CaO in the 5.0g sample and calculate it as a percentage of the total mass.

Let's start by calculating the amount of carbon dioxide (CO2) produced:

To liberate 1.32g of carbon(4) oxide (CO2), we know that the molecular weight of CO2 is 12 (C) + 16 (O) + 16 (O) = 44 g/mol.
Using the molar mass of CO2, we can calculate the number of moles of CO2 produced:
moles of CO2 = mass of CO2 / molar mass of CO2
moles of CO2 = 1.32 g / 44 g/mol = 0.03 moles of CO2

Next, we need to calculate the number of moles of CaO that would produce the same amount of CO2. From the balanced chemical equation, we know that 1 mole of CaO produces 1 mole of CO2:

Thus, the moles of CaO will also be 0.03 moles.

Now, let's calculate the amount of CaO in the 5.0g mixture:

From the periodic table, we know that the molecular weight of CaO is 40 (Ca) + 16 (O) = 56 g/mol.

Using the molar mass of CaO, we can calculate the mass of CaO:
mass of CaO = moles of CaO x molar mass of CaO
mass of CaO = 0.03 moles x 56 g/mol = 1.68 g

Finally, let's calculate the percentage of CaO in the mixture:
percentage of CaO = (mass of CaO / total mass of mixture) x 100
percentage of CaO = (1.68 g / 5.0 g) x 100 = 33.6%

Therefore, the percentage of CaO in the mixture is 33.6%.