The decomposition of KClO3(s) into KCl(s) and O2(g) with an 85% yield. How many moles of KClO3 must be used to generate 3.50 moles of O2?

Question

Naomi Aronson · Accepted Answer

2.75 moles

In this case, the reaction's equation is #2KClO_3(s) # -> #2KCl (s) + 3O_2(g) # Now as you see, the yield of the reaction is 85% which means that out of the number of moles of the products to be given out, only 80% is obtained. In order to obtain 3.50 moles of #O_2#, which is 85% of the no. of moles which ought to be obtained. So, if the reaction gave 100% of the yield, no. of moles of #O_2# would be 4.12 moles. (#0.85*x = 3.50 # ) Now, 3 molecules of #O_2# are given by 2 molecules of #KClO_3# No. of moles of 1 molecule of #O_2# = #4.12/3# By ratio, No. of moles of 2 molecules of #KCLO_3 # that reacted = #4.12*2/3# = 2.75 moles Therefore, 2.75 moles of #KCLO_3 # is required.

Cooper Anderson · Answer

undefined To generate 3.50 moles of O2 with an 85% yield, you would need to calculate the amount of KClO3 required based on the stoichiometry of the reaction. From the balanced chemical equation: 2 KClO3(s) → 2 KCl(s) + 3 O2(g), the molar ratio between KClO3 and O2 is 2:3.

So, if you have 3.50 moles of O2, you can set up a proportion to find the moles of KClO3 needed:

(3.50 moles O2 / 3 moles O2) * 2 moles KClO3 = 2.33 moles KClO3 (approximately)