Answer to Question #136131 in General Chemistry for Javier

Question

How many grams of O2 are needed in a reaction that produces 73.0 g Na2 O?

Answer 1

Let's

consider the reaction that produces Na₂O by the reaction of Na(s) and O₂(g) is...

4Na(s) + O₂(g) ------> 2Na₂O

Atomic mass of Na = 23 g/mole

Atomic mass of Oxygen = 16 g/ mole

Molicular mass of Oxygen (O₂) = 32 g/mole

Molecular mass of Na₂O = (2 × Atomic mass of Na + Atomic mass of Oxygen)

= ( 2 × 23 + 16 ) g/mole

= 62 g/mole

In the above reaction

2 moles of Na₂O produced by the reaction of 1 mole of molicular Oxygen (O₂)

1 mole of Na₂O = 62 g Na₂O

2mole of Na₂O = (62×2) g Na₂O

= 124 g Na₂O

So,

124 g Na₂O produced from 32 g O₂

1 g Na₂O produced from (32 g/124 g) O₂

73.0 g Na₂O produced from [(32 g/124 g) × 73.0 g] O₂ molicule

= 18.898 g O2 molicule

(Upto three significant decimal)

Hence,

18.898 grams of O₂ are needed in a reaction that produces 73.0 g Na₂O