calculate the number of moles of cacl2 that can be obtained from 25g of limestone caco3 in the prescence of excess hydrogen chloride

CaCO3 + 2HCl ==> CaCl2 + H2O + CO2

mols CaCO3 = grams CaCO3/molar mass CaCO3 = ?
mols CaCl2 = mols CaCO3 since 1 mol CaCO3 produces 1 mol CaCl2.
Then g CaCl2 = mols CaCl2 x molar mass CaCl2 = ?

111g = 1×6.03

0.4mol

answer

To determine the number of moles of CaCl2 that can be obtained from 25g of limestone (CaCO3) in the presence of excess hydrogen chloride (HCl), you need to follow a few steps.

Step 1: Write the balanced chemical equation for the reaction between limestone and hydrogen chloride:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2

This equation shows that one mole of CaCO3 reacts with two moles of HCl to produce one mole of CaCl2.

Step 2: Calculate the molar mass of CaCO3:
Molar mass of Ca = 40.08 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol

Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol)
= 40.08 g/mol + 12.01 g/mol + 48.00 g/mol
= 100.09 g/mol

Step 3: Calculate the number of moles of CaCO3:
Number of moles = mass / molar mass
= 25 g / 100.09 g/mol
= 0.2499 mol (rounded to 4 decimal places)

Step 4: Since the balanced equation shows that the mole ratio between CaCO3 and CaCl2 is 1:1, the number of moles of CaCl2 obtained will also be 0.2499 mol.

Therefore, the number of moles of CaCl2 that can be obtained from 25g of limestone (CaCO3) in the presence of excess hydrogen chloride is approximately 0.2499 mol.