Calculate the number of mole of cacl2 calcium chloride that can be obtained frm 30g of calcium trioxocarbonate iv in the presence of excess hydrochloride acid
The correct IUPAC name for that weird trioxo name is calcium carbonate.
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
How many mols CaCO3 do you have to start? That's mols = grams/molar mass = 30/100 = 0.30. How many mols CaCl2 will that form? That's 0.3 mols CaCO3 x (1 mol CaCl2/1 mol CaCO3) = 0.30 mols CaCl2.
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Hep me out guys
Well, first of all, let me just say that this sounds like a chemical party I wouldn't want to attend! But since you asked, let's do some mole calculations.
To find the number of moles of calcium trioxocarbonate IV (CaCO3) in 30g, we need to know its molar mass. The molar mass of CaCO3 is 100.09 g/mol.
So, the number of moles of CaCO3 is calculated using the formula:
moles = mass / molar mass
moles = 30g / 100.09 g/mol
moles ≈ 0.2999 mol
Now, to find the number of moles of CaCl2 formed, we need to know the balanced chemical equation between CaCO3 and HCl, which unfortunately you didn't provide. But assuming the reaction is as follows:
CaCO3 + 2HCl → CaCl2 + H2O + CO2
We can see that 1 mole of CaCO3 will produce 1 mole of CaCl2. Therefore, the number of moles of CaCl2 formed would be the same as the moles of CaCO3:
moles of CaCl2 ≈ 0.2999 mol
And there you have it! Approximately 0.2999 moles of CaCl2 can be obtained from 30g of CaCO3 in the presence of excess HCl. Please note that these calculations are based on assumptions, and the actual reaction and stoichiometry may differ.
To calculate the number of moles of calcium chloride (CaCl2) that can be obtained from 30g of calcium trioxocarbonate IV (CaCO3) in the presence of excess hydrochloric acid (HCl), we need to follow these steps:
Step 1: Determine the molar mass of CaCO3.
The molar mass of calcium trioxocarbonate IV (CaCO3) is calculated by adding up the atomic masses of each element. The atomic masses of calcium (Ca), carbon (C), and oxygen (O) are approximately 40.08 g/mol, 12.01 g/mol, and 16.00 g/mol, respectively. Hence, the molar mass of CaCO3 is:
Molar mass = (1 × Ca atomic mass) + (1 × C atomic mass) + (3 × O atomic mass)
Molar mass = (1 × 40.08 g/mol) + (1 × 12.01 g/mol) + (3 × 16.00 g/mol)
Molar mass of CaCO3 = 100.09 g/mol
Step 2: Calculate the number of moles of CaCO3.
To calculate the number of moles, we can use the formula: Moles = Mass / Molar mass
Number of moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3
Number of moles of CaCO3 = 30 g / 100.09 g/mol
Number of moles of CaCO3 = 0.2997 mol (approximately)
Step 3: Calculate the molar ratio between CaCO3 and CaCl2.
By examining the balanced chemical equation for the reaction between CaCO3 and HCl, we can determine the stoichiometric ratio between the two substances. The balanced chemical equation is:
CaCO3 + 2HCl → CaCl2 + H2O + CO2
From the equation, we see that 1 mole of CaCO3 reacts with 1 mole of CaCl2. Therefore, the molar ratio is 1:1.
Step 4: Calculate the number of moles of CaCl2.
Since the molar ratio is 1:1 between CaCO3 and CaCl2, the number of moles of CaCl2 is also 0.2997 mol.
Therefore, the number of moles of CaCl2 that can be obtained from 30g of CaCO3 is approximately 0.2997 mol.