Question

Soda ash (sodium carbonate) is widely used in the manufacture of glass. Prior to the environmental movement much of it was produced by the following reaction.
CaCO3 + 2 NaCl → Na2CO3 + CaCl2

Unfortunately, the byproduct calcium chloride is of little use and was dumped into rivers, creating a pollution problem. As a result of the environmental movement, all of these plants closed. Assume that 125g of calcium carbonate (100.09 g/mol) and 125 g of sodium chloride (58.44 g/mol) are allowed to react.

Determine how many grams of useful sodium carbonate (105.99 g/mol) will be produced.
How many grams of useless calcium chloride (110.98 g/mol) will be produced?
You should also determine how many grams of excess reactants are left (indicate which one is the limiting reactant)

Answer 1

Answer:

NaCl is the limiting reactant

18.0g of CaCO3 are in excess

113.4g Na2CO3 are produced

118.7g CaCl2 are produced

Explanation:

To solve this question we need to convert the mass of each reactant to moles in order to find the limitng and excess reactant to find the amount of products that would be produced:

Moles CaCO3:

125g * (1mol / 100.09g) = 1.25 moles

Moles NaCl:

125g * (1mol / 58.44g) = 2.14 moles

For a complete reaction of 2.14 moles of NaCl are required:

2.14 moles NaCl * (1mol CaCO3 / 2 mol NaCl) = 1.07 moles of CaCO3.

As there are 1.25 moles, Calcium carbonate is the excess reactant and NaCl the limiting reactant

The moles of CaCO3 in excess are:

1.25mol - 1.07mol = 0.18mol CaCO3 and the mass is:

0.18mol CaCO3 * (100.09g / mol) = 18.0g of CaCO3 are in excess

The moles of Na2CO3 and CaCl2 produced are:

2.14 moles NaCl * (1mol Na2CO3-1mol CaCl2 / 2 mol NaCl) = 1.07 moles are produced.

The masses are:

1.07 mol Na2CO3 * (105.99g / mol) = 113.4g Na2CO3

1.07 mol CaCl2 * (110.98g / mol) = 118.7g CaCl2