Question

1) Calculate the amount of 1.50 M Na2CO3 needed to react completely with 25.0 mL of 0.500 M CaCl2.

2) Calculate the theoretical yield of chalk (calcium carbonate) for each reaction (#1 & #2)—show all calculations. For reaction #1 CaCl2 should be limiting and for reaction #2 Na2CO3 should be limiting. You need to figure out how much sodium carbonate solution you will use in your experiment

Na₂CO₃(aq) + CaCl₂(aq) àCaCO₃(s) + 2NaCl(aq)  (molecular equation)

CO₃^2– (aq)  +  Ca²⁺(aq)  àCaCO₃(s)   (net ionic equation)

Limiting reactant(determining which is limiting and how much product you expect)

Step 1:  convert from grams of reactant “A” to moles of reactant “A”

Step 2:  convert moles of reactant “A” to moles of product (multiply by the ratio of elements from the balanced equation)

Step 3:  convert from moles of product to grams of product (this is the theoretical yield if reactant “A” is limiting)

Step 4:  convert from grams of reactant “B” to moles of reactant “B”

Step 5:  convert moles of reactant “B” to moles of product (multiply by the ratio of elements from the balanced equation)

Step 6:  convert from moles of product to grams of product (this is the theoretical yield if reactant “B” is limiting)

[repeat for any other reactants present]

Step 7:  the limiting reactant produces the least amount of product.

You only need one to predict the amount of CaCO₃that could be produced. Go ahead and evaluate that for each reactant independently. If both give the same amount of product, then neither is a limiting reagent. If there is a discrepancy, then the reagent that gives the smaller amount of product is the limiting one.  

Answer 1

1. The reaction is:

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

Moles of CaCl₂ used = Volume of the solution in Litres * Molarity of the solution = (25/1000) L * 0.500 M = 0.0125 mole

2. From the stoichiomtery , we get:

Moles of CaCO₃ produced = moles of Na₂CO₃ or CaCl₂ reacted = 0.0125 moles

Mass of CaCO₃ produced = 0.0125 * molar mass of CaCO₃ = 0.0125 mole * 100 g/ mole = 1.25 g

Therorectical yield of CaCO₃ is: 1.25 g

From the balanced stoichiometry, we get:

Moles of Na₂CO₃ / Moles of CaCl₂ = 1/1

So, moles of Na₂CO₃ reacted = Moles of CaCl₂ = 0.0125 mole

Mass of Na₂CO₃ reacted = 0.0125 mole * molar mass of Na2CO3 = 0.0125 mole * 106 g/ mole = 1.325 g

Volume of Na₂CO₃ solution required = Moles / Molarity of the solution = 0.0125 mole/ 1.50 M = 0.00833 L = 8.33 mL