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

CaCO3-> CaO + CO2

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To determine the number of grams of calcium carbonate required to prepare 50.0 grams of calcium oxide, we need to use the molar masses of the substances involved.

The molar mass of calcium carbonate (CaCO3) is calculated as follows:
1 atom of calcium (Ca) has a molar mass of 40.08 g/mol.
1 atom of carbon (C) has a molar mass of 12.01 g/mol.
3 atoms of oxygen (O) have a combined molar mass of (16.00 g/mol x 3) = 48.00 g/mol.

Adding these together:
40.08 g/mol + 12.01 g/mol + 48.00 g/mol = 100.09 g/mol.

Therefore, the molar mass of calcium carbonate is 100.09 g/mol.

The balanced equation shows that one mole of calcium carbonate produces one mole of calcium oxide:
CaCO3 -> CaO + CO2.

To find the amount of calcium carbonate needed, we set up a proportion using the molar mass of calcium carbonate and calcium oxide:

(50.0 g CaO) x (1 mol CaCO3/1 mol CaO) x (100.09 g CaCO3/1 mol CaCO3) = 50.0 g CaCO3

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

To determine the number of grams of calcium carbonate required to prepare 50.0 grams of calcium oxide, you need to consider the balanced chemical equation provided:

CaCO3 -> CaO + CO2

The coefficient in front of CaCO3 indicates the stoichiometric ratio between CaCO3 and CaO. In this case, the coefficient is 1, which means that 1 mole of CaCO3 is needed to produce 1 mole of CaO.

To calculate the number of moles of CaO, divide the given mass of CaO by its molar mass. The molar mass of CaO is determined by adding the atomic masses of calcium (Ca) and oxygen (O) together. The atomic mass of calcium is 40.08 g/mol, and the atomic mass of oxygen is 16.00 g/mol.

Molar mass of CaO = 40.08 g/mol + 16.00 g/mol = 56.08 g/mol

Moles of CaO = mass of CaO / molar mass of CaO
= 50.0 g / 56.08 g/mol

Now that we have the number of moles of CaO required, we can determine the number of moles of CaCO3 needed by using the stoichiometric ratio from the balanced chemical equation.

According to the balanced chemical equation, 1 mole of CaCO3 produces 1 mole of CaO. Therefore, the number of moles of CaCO3 needed is the same as the number of moles of CaO.

Therefore, the number of moles of CaCO3 needed is 50.0 g / 56.08 g/mol.

To find the mass of CaCO3 required, multiply the number of moles of CaCO3 by its molar mass.

Mass of CaCO3 = moles of CaCO3 * molar mass of CaCO3

Note: The molar mass of CaCO3 can be calculated by adding the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms together. The atomic mass of calcium is 40.08 g/mol, the atomic mass of carbon is 12.01 g/mol, and the atomic mass of oxygen is 16.00 g/mol.

Plug in the values and calculate:

Mass of CaCO3 = (50.0 g / 56.08 g/mol) * (40.08 g/mol + 12.01 g/mol + (3 * 16.00 g/mol))

Therefore, the mass of calcium carbonate required to prepare 50.0 grams of calcium oxide is equal to the calculated value using the above formula.

each mole of CaCO3 produces 1 mole of CaO

so, convert grams CaO to moles, then back to grams of CaCO3