Calculate the nu!her of moles of CaCL2 that can be obtained from 25g of lime stone (CaCo3) in the presence of external acid
To calculate the number of moles of CaCl2 that can be obtained from 25g of limestone (CaCO3) in the presence of an external acid, we need to consider the balanced chemical equation.
The balanced chemical equation for the reaction between limestone (CaCO3) and hydrochloric acid (HCl) is:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
From the equation, we can determine the molar ratio between CaCO3 and CaCl2, which is 1:1. This means that for every 1 mole of CaCO3, we should obtain 1 mole of CaCl2.
Now, we need to calculate the number of moles of CaCO3 in 25g using its molar mass.
The molar mass of CaCO3 is:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 oxygen atoms in CaCO3)
Molar mass of CaCO3 = 40.08 + 12.01 + (16.00 x 3) = 100.09 g/mol
Now, we can calculate the number of moles of CaCO3 in 25g using the formula:
moles = mass / molar mass
Number of moles of CaCO3 = 25g / 100.09 g/mol = 0.2499 mol (rounded to 4 decimal places)
Since the molar ratio between CaCO3 and CaCl2 is 1:1, the number of moles of CaCl2 obtained will be the same as the number of moles of CaCO3.
Therefore, the number of moles of CaCl2 that can be obtained from 25g of limestone (CaCO3) in the presence of an external acid is 0.2499 moles.