Question

Given the following:

4 KO₂ + 2 H₂O ===> 4 KOH + 3 O₂

Find the grams KO₂ and H₂O needed to make 1350 grams O₂

Answer 1

Given, The equilibrium reaction,

4KO₂ + 2H₂O 4KOH + 3O₂

Also given,

Mass of O₂ = 1350 grams

Now, calculating the number of moles of O₂,

= 1350 g of O₂ x ( 1 mol /31.998 g)

= 42.1901 mol O₂

Now, Using the moles of O₂ and the mole ratio from the balanced chemical reaction, calculating the number of moles KO₂ and

= 42.1901 mol O₂ x ( 4 mol KO₂ / 3 mol O₂)

= 56.2535 mol KO₂

Converting the number of moles to grams,

= 56.2535 mol KO₂ x (71.1 g / 1 mol)

= 3999.6 g of KO₂ Or 4.00 x 10³ g of KO₂ [ 3 S.F]

Similarly,

= 42.1901 mol O₂ x ( 2 H₂O / 3 mol O₂)

= 28.1268 mol H₂O

Converting the number of moles to grams,

= 28.1268 mol H₂O x (18.02 g / 1 mol)

= 506.8 g of H₂O Or 507 g of H₂O [ 3S.F]