how many grams of na2co3 are there are?

Question

When 35.00 mL of Na₂CO₃ was reacted with 29.06 mL of a 0.125 M CaCl₂ solution, all of the carbonate ion precipitated as CaCO₃. How many grams of Na₂CO₃ were in the solution?

Answer 1

Write the correctly balanced equation:

Na₂CO₃(aq) + CaCl₂(aq) ==> CaCO₃(s) + 2NaCl(aq)

Moles of CaCl₂ present = 0.125 mol/L x 0.02906 L = 0.003633 moles

Moles Na₂CO₃ present = 0.003633 mol CaCl₂ x 1 mol Na₂CO₃/1 mol CaCl₂ = 0.003633 moles

Grams Na₂CO₃ present = 0.003633 moles x 105.989 g/mole = 0.385 g Na₂CO₃ (to 3 significant figures)