Question

Consider the oxidation of sodium metal to sodium oxide described by the balanced equation:

Answer 1

Answer:

1116 g.

Explanation:

The balanced equation for the reaction is given below:

4Na + O₂ —> 2Na₂O

From the balanced equation above,

1 mole of O₂ reacted to produce 2 moles of Na₂O.

Next, we shall determine the theoretical yield of Na₂O. This can be obtained as follow:

From the balanced equation above,

1 mole of O₂ reacted to produce 2 moles of Na₂O.

Therefore, 9 moles of O₂ will react to produce = 9 × 2 = 18 moles of Na₂O.

Finally, we shall determine the mass in 18 moles of Na₂O. This can be obtained as follow:

Mole of Na₂O = 18 moles

Molar mass of Na₂O = (23×2) + 16

= 46 + 16

= 62 g/mol

Mass of Na₂O =?

Mass = mole × molar mass

Mass of Na₂O = 18 × 62

Mass of Na₂O = 1116 g

Thus, the theoretical yield of Na₂O is 1116 g.