Answer to Question #90902 in General Chemistry for acadia carlson

Question

Calculate the number of grams of sodium peroxide, Na2O2, that would be required to produce 50 g of oxygen by the following reaction: 2 Na2O2 + 2 H2O  4 NaOH + O2
How many litres of pure O2 at STP, are required to react completely with 2.43 g of magnesium?

Answer 1

Solution: n=m/M, 50/32=1,5625 (50 g of oxygen is 1,5625 mol)

Producing 1 mol of oxygen by the reaction between sodium peroxide and .water requires 2 moles of Na2O2, so for 1,5625 moles of oxygen you need 1,5625*2=3,125 moles of Na2O2. The mass of it is 3,125*78=243,75 grams.

The reaction between magnesium and oxygen is 2Mg+O2=2MgO. 2,43 g of magnesium is 2,43/24,3=0,1 mol, so amount of oxygen needed for the reactionn is 0.1/2=0,05 mol. At STP 1 mol of any gas has a volume of 22,4 liters, so 0,05 moles will take 22,4*0,05=1,12 liters.

Answer: m(Na2O2)=243,75 g, V(O2)=1,12 l.