Final answer:
Approximately 0.1 moles of calcium trioxocarbonate (IV) were used in the preparation by converting 10 g of CaCO3 to moles using its molar mass of 100.09 g/mol.
Step-by-step explanation:
To calculate the amount of calcium trioxocarbonate (IV), also known as calcium carbonate (CaCO3), used in the preparation of carbon (IV) oxide, we need to determine the molar mass of CaCO3 and use it to convert grams to moles.
The molar mass of CaCO3 is found by summing the atomic masses of calcium (Ca), carbon (C), and oxygen (O), which are approximately 40.08 g/mol, 12.01 g/mol, and 16.00 g/mol respectively. Given that CaCO3 has one atom of calcium, one atom of carbon, and three atoms of oxygen, the molar mass of CaCO3 is:
(1 × 40.08 g/mol) + (1 × 12.01 g/mol) + (3 × 16.00 g/mol) = 100.09 g/mol.
Now, we can calculate the number of moles of CaCO3 using the mass provided and the molar mass:
Number of moles = Mass in grams / Molar mass
= 10 g / 100.09 g/mol
Approximately 0.1 moles of calcium trioxocarbonate (IV) were used in the preparation.