Question

What is the theoretical yield of sodium oxide (Na₂O) in grams when 20.0g of calcium oxide (CaO) reacts with an excess amount of Sodium chloride? *
10 points
Captionless Image
10.0g
22.1g
32.9g
5.6g

Answer 1

Answer:

22.1g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CaO + 2NaCl —> Na2O + CaCl2

Next, we shall determine the mass of CaO that reacted and the mass of Na2O produced from the balanced equation.

This is illustrated below:

Molar mass of CaO = 40 + 16 = 56g/mol

Mass of CaO from the balanced equation = 1 x 56 = 56g

Molar mass of Na2O = (23x2) + 16 = 62g/mol

Mass of Na2O from the balanced equation = 1 x 62 = 62g

From the balanced equation above,

56g of CaO reacted to produce 62g of Na2O.

Finally, we can determine the theoretical yield of Na2O as follow:

From the balanced equation above,

56g of CaO reacted to produce 62g of Na2O.

Therefore, 20g of CaO will react to produce = (20 x 62)/56 = 22.1g of Na2O.

Therefore, the theoretical yield of Na2O is 22.1g

Answer 2

Answer:

22.1 g

Explanation:

The balanced reaction equation which serves as a guide in solving the problem is given as;

CaO(s) + 2NaCl(aq) ------> Na2O(s) + CaCl2(aq)

The question clearly specifies that sodium chloride is the reactant in excess. This means that calcium oxide should be used to calculate the theoretical yield of sodium oxide.

Number of moles of calcium oxide reacted = mass of calcium oxide / molar mass of calcium oxide

Molar mass of calcium oxide = 56.0774 g/mol

Mass of calcium oxide = 20.0g

Number of moles of calcium oxide = 20.0 g/ 56.0774 g/mol = 0.3566 moles

From the balanced reaction equation;

1 mole of calcium oxide produces 1 mole of sodium oxide

Therefore, 0.3566 moles of calcium oxide will produce 0.3566 moles of sodium oxide.

Mass of sodium oxide produced = number of moles × molar mass

Molar mass of sodium oxide= 61.9789 g/mol

Mass of sodium oxide = 0.3566 moles × 61.9789 g/mol

Mass of sodium oxide= 22.1 g