Question

When a mixture of 10 moles of $ S{O_2} $ and 15 moles of $ {O_2} $ was passed over catalyst, 8 moles of $ {\text{S}}{{\text{O}}_{\text{3}}} $ was formed. How many moles of $ S{O_2} $ and $ {O_2} $ did not enter into combination?

Answer 1

Hint: In a chemical process, the limiting reagent (or limiting reactant or limiting agent) is a reactant that is completely consumed when the reaction is finished. This reagent limits the amount of product produced since the reaction cannot continue without it. Surplus reagents or excess reactants occur when one or more additional reagents are present in proportions greater than those necessary to react with the limiting reagent (xs).

Complete answer:
Because the theoretical yield is defined as the quantity of product produced when the limiting reagent reacts fully, the limiting reagent must be determined in order to compute the percentage yield of a reaction. There are numerous comparable approaches to determine the limiting reagent and assess the surplus amounts of other reagents given the balanced chemical equation that describes the reaction.
S=32.06u molar mass
O has a molar mass of 16 u.
 $ S{O_2} $ has a molar mass of 64.06u.
10 mole of $ S{O_2} $ mass in grams=10 x 32.06 = 320.6g
15 mole mass in grams of $ {O_2} $ 15 x 32.00 = 960g of $ {O_2} $
The mass of S that was injected was 10 x 32.06 = 320.6g.
The mass of S in the form of $ S{O_3} $ is 8 x 32.06 = 256.48g.
Moles of S that did not participate in the reaction is 320.6 - 256.48 = 64.12g = 2 moles of S
Now, because S participated in the reaction in the form of $ S{O_2} $ , if there were 2 moles of S that did not participate in the reaction, there were also 2 moles of $ S{O_2} $ .
When $ S{O_2} $ is removed from $ S{O_3} $ , all that is left is O, thus the amount of $ {O_2} $ consumed in the process is 8(number of moles of O in $ S{O_3} $ ) x 16.00(molar mass of oxygen) =128g.
15 x 32.00(molar mass of $ {O_2} $ ) = 480.00g is the mass of oxygen that went in.
As a result, the mass of oxygen not utilised was = 480.00 - 128.00 = 352g = 11 moles of oxygen.
As a result, the reaction did not include 2 moles of $ S{O_2} $ and 11 moles of $ {O_2} $ .

Note:
When one of the reactants of a chemical reaction is depleted, the process abruptly comes to a halt. It must be established which reactant will restrict the chemical reaction (the limiting reagent) and which reactant is in excess to determine the amount of product generated (the excess reagent). Calculating the quantity of product that each reactant may generate is one approach to discover the limiting reagent; the one that produces the least product is the limiting reagent.