Limited Reactants?

Question

How many grams SO2 can be formed from 20.0 g of S and 160g O2?


How many gramsSO2 can be formed from 20 g S and 15.0 g of O2?

Answer 1

First, set up the balanced chemical reaction.

S + O2 ----> SO2

which shows one mole of S will react with one mole of O2 gas to form 1 mole of SO2 gas.

Next, convert the respective grams into moles.
20.0g S / MM S = 20.0/ 32 = 0.625 moles

160 g O2/MM O2 = 160/16x2 = 160/32 = 5
clearly, there is a lot of moles of Oxygen than S, therefore the limiting reagent, the one that is lesser, is S.

Then, this means that the number of moles of SO2 is limited by the amount of initial amount of S available to begin with, which is .625 moles. Note the one to one ratio in the balanced eqn.

Finally, to get the grams of SO2 multiply the .625 moles by the MM of SO2. ( 32+ 32)

Or, 0.625 x 64 = 40 grams.

In the other example the limiting reagent is now O2 not S. I will leave you to do the calculations. You need the practice. Good luck.

You have to analyze. Not memorize.

In any case, you MUST have a BALANCED equation, else all numbers will be wrong even if you identified the limiting reagent. Good Luck.

Answer 2

S + O2 -> SO2

40 g

30 g