To determine how many grams of MgC03 contain 24.00 g of oxygen, we need to use the molar mass of MgC03 and convert the given mass of oxygen into moles.

The molar mass of MgC03 is given as 84.30 g/mol, which means that one mole of MgC03 weighs 84.30 grams.

To convert the mass of oxygen into moles, we divide the given mass by the molar mass of oxygen. The molar mass of oxygen is 16.00 g/mol.

24.00 g of oxygen / 16.00 g/mol = 1.50 moles of oxygen.

Now, we need to find out the mass of MgC03 that contains 1.50 moles of oxygen.

Since the molecular formula of MgC03 contains one atom of oxygen, and the molar mass of MgC03 is 84.30 g/mol, we can calculate the mass of MgC03 as follows:

1.50 moles of oxygen * 84.30 g/mol = 126.45 g of MgC03.

Therefore, the exact amount of MgC03 needed to contain 24.00 g of oxygen is 126.45 g.

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