<u>Given:</u>
Volume of Na2CO3 = 250 ml = 0.250 L
Molarity of Na2CO3 = 6.0 M
Volume of CaF2 = 750 ml = 0.750 L
Molarity of CaF2 = 1.0 M
<u>To determine:</u>
The mass of CaCO3 produced
<u>Explanation:</u>
Na2CO3 + CaF2 → CaCO3 + 2NaF
Based on the reaction stoichiometry:
1 mole of Na2CO3 reacts with 1 moles of Caf2 to produce 1 mole of caco3
Moles of Na2CO3 present = V * M = 0.250 L * 6.0 moles/L = 1.5 moles
Moles of CaF2 present = V* M = 0.750 * 1 = 0.750 moles
CaF2 is the limiting reagent
Thus, # moles of CaCO3 produced = 0.750 moles
Molar mass of CaCO3 = 100 g/mol
Mass of CaCO3 produced = 0.750 moles * 100 g/mol = 75 g
Ans: Mass of CaCO3 produced = 75 g
