5.00gram of a mixture of caco3 and cao liberated 1.32 gram of co2 on strong heating what is the percentage of caco3 in the mixture
The percentage of CaCO3 in the mixture is 75.2%. This is calculated by dividing the mass of CaCO3 (3.68 g) by the total mass of the mixture (5.00 g), and then multiplying by 100.
To find the percentage of CaCO3 in the mixture, we need to first calculate the mass of CaCO3 that was present in the mixture.
Given:
Mass of the mixture = 5.00 grams
Mass of CO2 liberated = 1.32 grams
We know that when CaCO3 is heated, it decomposes to form CO2 and CaO according to the following balanced chemical equation:
CaCO3 --> CaO + CO2
From the equation, we can see that for every 1 mole of CaCO3, 1 mole of CO2 is produced.
To find the number of moles of CO2 liberated, we need to use the molar mass of CO2, which is 44.01 g/mol.
Number of moles of CO2 = mass of CO2 / molar mass of CO2
= 1.32 / 44.01
≈ 0.03 mol (rounded to two decimal places)
Since the number of moles of CaCO3 and CO2 are the same, the number of moles of CaCO3 in the mixture is also 0.03 mol.
Now, to find the mass of CaCO3, we can use the molar mass of CaCO3, which is 100.09 g/mol.
Mass of CaCO3 = number of moles of CaCO3 × molar mass of CaCO3
= 0.03 × 100.09
= 3.00 grams
Therefore, the mass of CaCO3 in the mixture is 3.00 grams.
Now, we can calculate the percentage of CaCO3 in the mixture:
Percentage of CaCO3 = (mass of CaCO3 / mass of the mixture) × 100
= (3.00 / 5.00) × 100
≈ 60% (rounded to the nearest whole number)
So, the percentage of CaCO3 in the mixture is approximately 60%.
To find the percentage of CaCO3 in the mixture, we need to calculate the mass of CaCO3 in the mixture first.
Given:
Mass of mixture = 5.00 grams
Mass of CO2 liberated = 1.32 grams
First, we need to determine the molar mass of CO2. The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol. Since there are two oxygen atoms in CO2, we get:
Molar mass of CO2 = 12.01 g/mol + (16.00 g/mol x 2) = 44.01 g/mol
Now, let's calculate the moles of CO2 liberated:
Moles of CO2 = Mass of CO2 / Molar mass of CO2
Moles of CO2 = 1.32 g / 44.01 g/mol = 0.03 moles
Since CaCO3 reacts with CO2 in a 1:1 ratio, the moles of CaCO3 is also 0.03 moles.
The molar mass of CaCO3 is:
Molar mass of CaCO3 = (40.08 g/mol + 12.01 g/mol + (16.00 g/mol x 3)) = 100.09 g/mol
Now, we can calculate the mass of CaCO3 in the mixture:
Mass of CaCO3 = Moles of CaCO3 * Molar mass of CaCO3
Mass of CaCO3 = 0.03 moles * 100.09 g/mol = 3.00 grams
Finally, to find the percentage of CaCO3 in the mixture, we divide the mass of CaCO3 by the mass of the mixture and multiply by 100:
Percentage of CaCO3 = (Mass of CaCO3 / Mass of mixture) * 100
Percentage of CaCO3 = (3.00 g / 5.00 g) * 100 = 60%
Therefore, the percentage of CaCO3 in the mixture is 60%.