Determine the number of moles of Na2CO3 in the original 50.0 mL of solution

Question

The student filters and dries the precipitate of CaCO₃ (molar mass 100.1 g/mol) and records the data in the table below.

Volume of Na2CO3 solution	50.0 mL
Volume of 1.0 M Ca(NO3)2 added	100.0 mL
Mass of CaCO3 precipitate collected	0.93 g

 

Determine the number of moles of Na₂CO₃ in the original 50.0 mL of solution

Answer 1

First thing is to write the balanced equation for what is happening:

Na₂CO₃(aq) + Ca(NO₃)₂(aq) ==> CaCO₃(s) + 2NaNO₃(aq) ... balanced equation

moles Na₂CO₃ present = ?

moles Ca(NO₃)₂ present = 100.0 ml x 1 L/1000 ml x 1.0 mol/L = 0.100 moles Ca(NO₃)₂

moles CaCO₃ formed = 0.93 g x 1 mol/100.1 g = 0.00929 moles CaCO₃

From the balanced equation we see that Na₂CO₃ and Ca(NO₃)₂ react in a 1:1 mole ratio producing CaCO₃ that is also in a 1:1 mole ratio. Using the fact that 1 mole CaCO₃ comes from 1 mole of Na₂CO₃, we can determine moles of Na₂CO₃ originally present as follows:

0.00929 moles CaCO₃ x 1 mol Na₂CO₃ / 1 mol CaCO₃ = 0.00929 moles Na₂CO₃ originally present

Note: the fact that this value is significantly less than the moles of Ca(NO₃)₂ added tells us that the Ca(NO₃)₂ was not a limiting reagent.