Calculate the number of moles of calcium chloride that can be obtained in 30g of calcium trioxocarbonate (vi) in the presence of excess hydrochloric acid

Where did you get the name of calcium trioxocarbonate (vi)? There is no such compound. You may have meant CaCO3 which is properly called calcium carbonate.

CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
mols CaCO3 = g/molar mass = about 30/100 = 0.30
So you can 0.30 mols CaCl2 from 0.30 mols CaCO3.

To calculate the number of moles of calcium chloride that can be obtained from 30g of calcium trioxocarbonate (VI) in the presence of excess hydrochloric acid, we need to use stoichiometry.

The balanced chemical equation for the reaction between calcium trioxocarbonate (VI) (CaCO3) and hydrochloric acid (HCl) is:

CaCO3 + 2HCl → CaCl2 + H2O + CO2

From the equation, we can see that one mole of CaCO3 reacts with two moles of HCl to produce one mole of CaCl2.

To find the number of moles of CaCl2 produced, we can follow these steps:

1. Calculate the molar mass of CaCO3:
- The atomic mass of calcium (Ca) = 40.08 g/mol
- The atomic mass of carbon (C) = 12.01 g/mol
- The atomic mass of oxygen (O) = 16.00 g/mol (there are three oxygen atoms in CaCO3)
- Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (16.00 g/mol x 3) = 100.09 g/mol

2. Convert the given mass of CaCO3 to moles:
- Moles of CaCO3 = (mass of CaCO3) / (molar mass of CaCO3)
- Moles of CaCO3 = 30 g / 100.09 g/mol = 0.2998 mol (rounded to 4 decimal places)

3. According to the balanced equation, the stoichiometric ratio between CaCO3 and CaCl2 is 1:1.
Therefore, the number of moles of CaCl2 formed will be equal to the number of moles of CaCO3.

Therefore, the number of moles of calcium chloride that can be obtained from 30g of calcium trioxocarbonate (VI) in the presence of excess hydrochloric acid is approximately 0.2998 moles.

To calculate the number of moles of calcium chloride obtained, you need to consider the balanced chemical equation for the reaction between calcium trioxocarbonate (VI) and hydrochloric acid.

The chemical equation for the reaction is:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2

From the equation, you can see that one mole of calcium trioxocarbonate reacts with two moles of hydrochloric acid to produce one mole of calcium chloride.

First, find the molar mass of calcium trioxocarbonate (VI) or CaCO3:
The atomic masses are: Ca - 40.08 g/mol, C - 12.01 g/mol, and O - 16.00 g/mol.
Adding these atomic masses together:
40.08 + 12.01 + (16.00 * 3) = 100.09 g/mol

Next, calculate the number of moles of calcium trioxocarbonate (VI) in 30g:
Moles = Mass / Molar mass
Moles = 30g / 100.09 g/mol = 0.2999 mol (approximately 0.30 mol)

Since the reaction between calcium trioxocarbonate (VI) and hydrochloric acid is 1:2, the moles of calcium chloride formed will also be 0.30 mol.

Therefore, the number of moles of calcium chloride obtained in 30g of calcium trioxocarbonate (VI) is 0.30 mol.