Oxygen is produced by the reaction of sodium peroxide and water.

2Na2O2 (s) + 2H2O (l) - - - > O2 (g) + 4NaOH (aq)

a. Calculate the mass of Na2O2 in grams needed to form 4.80g of oxygen.

b. How many grams of NaOH are produced when 4.80 g of O2 is formed?

c. When 0.48g of Na2O2 is dropped in water, how many grams of O2 are formed?

First you should start by calculating the molar mass of each compounds and elements so molar mass for Na2O2 = 77.98g, O2=16.0g and NaOH = 39.997g now we start with setting up the problems

a) 4.80g O2 (1mol O2/16g o2) (2 mol Na2O2/1mol O2) (77.98g NaO2/zmol NaO2) = 46.8 g of Na2O2 are required to form 4.80g of oxygen.

b) 4.80g O2 (1mol O2/16g O2) (4 mol NaOH/1 mol O2) (33.997g naOH/1 mol NaOH) = 48.0 g are produced when 4.80g of O2 is formed

c) 0.48g Na2O2 (1mol Na2O2/77.98g Na2O2) (1 mol O2/2 mol Na2O2) (16.0g O2 / 1 mol O2) = 0.049g of O2 are formed.