1.64 moles of ALC L₃ are produced upon the reaction of 4.5 moles of ALT and 10.5 moles of HCl with a 36.5% yield.
How to calculate the amount?
Calculate the moles of each reactant based on their given amounts:
Moles of ALT = 4.5 mol
Moles of HCl = 10.5 mol
Compare the moles of each reactant to their stoichiometric coefficients in the balanced equation (assuming the reaction is 1:1:1):
ALT: 1 mol ALT reacts with 1 mol HCl and produces 1 mol ALC L₃
HCl: 1 mol HCl reacts with 1 mol ALT and produces 1 mol ALC L₃
In this case, ALT (4.5 mol) is the limiting reagent because it is present in a smaller amount compared to the stoichiometric requirement of 6.75 mol (1.5 times the amount of HCl).
Since ALT is the limiting reagent, it will completely react and determine the maximum amount of ALC L₃ that can be produced.
Therefore, the theoretical moles of ALC L₃ produced is equal to the moles of ALT: 4.5 mol
The given yield (36.5%) indicates the percentage of the theoretical yield that is actually obtained.
To find the actual amount of ALC L₃, multiply the theoretical moles by the yield:
Actual moles of ALC L₃ = 4.5 mol × 0.365
= 1.64 mol
Therefore, 1.64 moles of ALC L₃ are produced upon the reaction of 4.5 moles of ALT and 10.5 moles of HCl with a 36.5% yield.
Complete question:
How much ALC L3 is produced upon reaction of 4.5 moles of ALT 10.5 moles of HCL with a 36.5% yield?