How much oxygen (in grams) will form from 31.2 kg of N2O5?

Answer 1

If you’re referring to this reaction?

2N2O5 ——- 4NO2 + O2

you take these steps a) convert Kg of N2O5

to grams, then convert the grams to moles of N2O5.

b) using molar ratios calculate the number of moles of O2 produced and then convert that to grams for your answer. Here are the steps in detail.

a) 31.2 Kg of N2O5 x 1000 g/1 Kg= 31200 g

now convert the grams of N2O5 to moles

molar mass of N2O5 = 2 N 28 g + 5 O 80 g =

108 g

31200 g x 1 mole/108 g = 289 moles N2O5

b) now use the molar ratios between N2O5 and O2.

if you look at the coefficients in front of each you see a 2 in front N2O5 meaning 2 moles and in front of the product O2 you see no number which implies 1 mole.

This tells you in this reaction 2 moles of N2O5 are needed to produce 1 mole of O2. So the molar ratio is 2:1.

here’s how you would use this:

289 mol N2O5 x 1 mol O2/2 mol N2O5

the N2O5 cancels and your left with moles of O2, you do the math calculations there and you get

144.5 moles of O2 gas produced.

last step convert the 144.5 moles of O2 to grams for your answer.

molar mass of O2 = 32 g

144.5 moles O2 x 32 g / 1 mole = 4624 grams of O2 gas

if this is not the reaction you are referring to as you didn’t in your question give a complete formula to work with message me with the full reaction you want to deal with and I’ll be happy to show you the steps again for that reaction.