Determine the grams of CaCO3 formed from 25.00mL of 3.50M Na2CO3 and excess CaCl2

Question

Question

Answer 1

Answer #1

Answer

8.758g

Explanation

The balanced equation for reaction between Na2CO3 and CaCl2 is as follows

Na2CO3 + CaCl2 -------> 2NaCl + CaCO3

This is 1:1 molar reaction

stoichiometrically, 1mole of Na2CO3 react with 1mole of CaCO3

Molarity = number of moles of solute per liter of solution

given moles of Na2CO3 =(3.50mol/1000ml)× 25ml = 0.0875mol

moles of CaCO3 formed = 0.0875mol

mass = Number of moles × molar mass

mass of CaCO3 formed = 0.0875mol × 100.09g/mol = 8.758g

Answer 2

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