What mass of calcium carbonate is produced when 250 ml of 6.0 m sodium carbonate is added to 750 ml of 1.0 m calcium fluoride

Given:

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

To determine:

The mass of CaCO3 produced

Explanation:

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

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.