Chemistry HW

Question

The metathesis reaction between sodium carbonate and cobalt (III) chloride is shown below

3 Na2CO3 (aq) + 2 CoCl3 (aq) ➝ Co2(CO3)3 (s) + 6 NaCl (aq)

How many grams of solid cobalt (III) carbonate can be produced by the reaction of 250.0 ml of 0.325 M sodium carbonate solution with 175.0 ml of 0.800 M cobalt (III) chloride solution?

Answer 1

Hello Jana, 

 

Look at the coefficients of the balanced equation:  

 

 

      3 Na2CO3 (aq)    +   2 CoCl3 (aq) ➝    Co2(CO3)3 (s)    +   6 NaCl (aq)

 

mole ratio:       3 mol                     2mol                     1mol                        6 mol

 

 

To find the amount of cobalt (III) carbonate that will be produced by mixing 250.0 ml of 0.325 M sodium carbonate solution with 175.0 ml of 0.800 M cobalt (III) chloride solution, you have to find out which reactant controls the amount of products formed (i.e. the limiting reactant). 

 

The simplest approach will be find the amount of cobalt (III) carbonate formed from each reactant, and then the reactant producing less amount of product is the limiting.

 

 

250.0 ml of 0.325 M Na₂CO₃  = 0.2500 L x 0.325 mol/L = 0.0813 mol Na₂CO₃

 

175.0 ml of 0.800 M CoCl₃    =   0.1750 L x 0.800 mol/L  =  0.14 mol CoCl₃

 

 

Consider 0.0813 mol Na₂CO₃: 

 

mol of Co₂(CO₃)₃  formed =  (1 mol Co₂(CO₃)₃/ 3 mol Na₂CO₃) x 0.0813 Na2CO3 = 0.0271 mol Co₂(CO₃)₃

 

Consider 0.14 mol CoCl₃:

 

mol of Co₂(CO₃)₃ formed   = (1 mol Co₂(CO₃)₃ /2 mol CoCl₃ ) x 0.14 mol CoCl₃   = 0.070 mol Co₂(CO₃)₃ 

 

 

since 0.070 is less than 0.0271, CoCl₃ controls the product and 0.070 mol Co₂₍Co₃₎₃ will be produced. 

 

 

Using the molar mass of Co₂(CO₃)₃ = 297.89 g/mol

 

0.070 mol Co₂(CO₃)₃ x 297.89 g/mol =  20.9 g Co₂(CO₃)₃ will be produced.