Calculate the number of moles of cacl2 that can be obtained from 25g of lime stone(caco3) in the presence of excess hydrogen chlorine
find moles of limestone ... divide mass (25g) by molar mass of CaCO3
since CaCl2 and CaCO3 both contain one mole of Ca per mole of compound
... you have your answer
molar mass of CaCl2=111g. To get the number of moles of CaCl2, divide the reacting mass by the molar mass that is, 25g/111g
How did you get 111g
To calculate the number of moles of CaCl2 that can be obtained from 25g of limestone (CaCO3), we need to follow these steps:
Step 1: Determine the molar mass of CaCO3.
The molar mass of an element or compound is the mass of one mole of the substance. To calculate the molar mass of CaCO3, we need to sum up the atomic masses of each element in one molecule of CaCO3.
The atomic masses are as follows:
- Ca (Calcium) = 40.08 g/mol
- C (Carbon) = 12.01 g/mol
- O (Oxygen) = 16.00 g/mol (There are three oxygen atoms in CaCO3)
Using these values, we can calculate the molar mass of CaCO3:
Molar mass of CaCO3 = (1 * Ca) + (1 * C) + (3 * O)
= (1 * 40.08) + (1 * 12.01) + (3 * 16.00)
= 40.08 + 12.01 + 48.00
= 100.09 g/mol
Step 2: Convert the given mass of limestone to moles.
Now that we have the molar mass of CaCO3, we can convert the given mass (25g) to moles using the formula:
Number of moles = Mass / Molar mass
Number of moles of CaCO3 = 25g / 100.09 g/mol
≈ 0.2498 mol (rounded to four decimal places)
Step 3: Determine the Stoichiometry.
The chemical equation for the reaction between CaCO3 and HCl can be written as:
CaCO3 + 2HCl → CaCl2 + H2O + CO2
Based on the balanced equation, we can see that one mole of CaCO3 reacts with two moles of HCl to produce one mole of CaCl2. Therefore, the number of moles of CaCl2 produced will be the same as the number of moles of CaCO3 used.
Number of moles of CaCl2 = 0.2498 mol
So, the number of moles of CaCl2 that can be obtained from 25g of limestone (CaCO3) in the presence of excess hydrogen chloride is approximately 0.2498 moles.