According to the explanation given in our previous session, now we have a similar question but with a slight difference, but first let's set up the reaction:
2 RbNO3 -> 2 RbNO2 + O2
We have 5.0 moles of RbNO3
From the molar ratio we see that 2 moles of RbNO3 is equal to 1 mol of O2, therefore if we have 5 moles of RbNO3, we will have 2.5 moles of O2 being produced.
Since the question is asking "how many liters" we have to assume that we are dealing with gases and these gases are at STP (standard temperature and pressure), which is T = 273K and P = 1 atm, in these specfici conditions, 1 mol of gases will have a volume of 22.4 Liters, therefore if O2 has 2.5 moles of O2
1 mol = 22.4L
2.5 moles = x L
x = 56 Liters of O2 are being produced with 5.0 moles of RbNO3