Question

Calculations (attach extra page as needed) Assuming 100% reaction, how many millimoles of calcium carbonate can be formed in each of the six solutions? Include calculations of millimoles of the starting amounts of calcium chloride and sodium carbonate solutions. d
Based upon the following given information:

Not sure of the type of equation needed for this question. The labs are ahead of the lectures so I am teaching myself as I go along.

Answer 1

First and foremost, we need to know the reaction involved, which is:

$Na_{2}CO_{3}+CaCl_{2}\rightarrow CaCO_{3}+2NaCl$

Which is telling us that sodium carbonate and calcium chloride react in a 1:1 ratio. Let's take a look at the reactions.

1) The amount of millimoles can be calculated by multiplying the molar concentration of the reagent solutions by the volume in units of mL. (Remember that molar concentration is equivalent to moles per liter or, at the same time, to mmoles per mL). In this case, since both reagents are in the same concentration:

$n=1\frac{mmol}{mL}\cdot 5mL=5mmoles$

We have 5 mmoles of each reactant, and we know that they combine in a 1:1 fashion to yield 1 moles of product, So from combining 5 mmoles of each reactant, 5 mmoles of CaCO₃ will be obtained.

2) Same procedure. We have 1 M Na₂CO₃ and 0.5 M CaCl₂. The number of mmoles of carbonate is the same as in the previous item (5 mmoles) and for CaCl₂:

$n=0.5\frac{mmol}{mL}\cdot 5mL=2.5mmoles$

Since there is a smaller number of moles of CaCl₂, this reactant is the limiting reactant. This means that only 2.5 mmoles of sodium carbonate can react with the 2.5 mmoles of calcium chloride present, so we will have 2.5 mmoles of CaCO₃ as product (there will also be 2.5 spare mmoles of Na₂CO₃).

3) This is similar to item 1, but with 2.5 mmoles of reagents to begin with, so, analogously proceeding, the amount of product will be 2.5 mmoles.

4) This is a case similar to that of item 2, where sodium carbonate is now the limiting reactant, so we will obtain 2.5 mmoles of product and there will be an excess of 2.5 mmoles of calcium chloride that will not react.

5) In this case, there is 0.1 M CaCl₂, which represents a number of mmoles of:

$n=0.1\frac{mmol}{mL}\cdot 5mL=0.5mmoles$

Since there are 2.5 mmoles of Na₂CO₃, calcium chloride will be the limiting reactant and 0.5 mmoles of product will be formed, while 2 mmoles of sodium carbonate will not react.

6) We now have 0.05 M CaCl₂, so the number of mmoles in 5 mL is:

$n=0.05\frac{mmol}{mL}\cdot 5mL=0.25mmoles$

There are, again 2.5 mmoles of sodium carbonate, so the limiting reactant is calcium chloride and we will obtain 0.25 mmoles of product, while 2.25 mmoles of Na₂CO₃ will not react.