Question

Potassium bromide KBr contains 32.9% by mass potassium, if 6.40 g of bromine reacts with 3.60 g of potassium, calculate the number of moles of potassium which combine with bromine to form KBr.

Answer 1

Hint: The general knowledge of limiting reactant, excess reactant and knowing the balanced equation for the above-mentioned reaction can help us solve the given illustration.
Limiting reactant always decides the extent of reaction taking place. So, all the calculation will be done by considering limiting reactant as the basis.

Complete answer:
Let us know about the terms involved in the given problem;
Limiting reactant-
The reactant which is consumed totally when the reaction is completed, is known as limiting reactant. The amount of product formed will be controlled by this reactant as scarcity of this will result in no further reaction.
Excess reactant-
The reactant which would remain in traces with the product when the reaction is completed i.e. limiting reactant is consumed. This would either be wasted or be recovered as there is no limiting left that would react with excess reactant.
Keeping this information in mind, let us move forward;
Illustration-
Given data-
Potassium bromide KBr contains 32.9% by mass potassium i.e. 100 g of KBr will have 32.9 g of K.
Thus, it has 67.1 g of bromine.
The reaction can be stated as,
\[2K+B{{r}_{2}}\to 2KBr\]
We know that,
67.1 g of bromine will react with 32.9 g of potassium.
Thus,
6.40 g of bromine will react with $\dfrac{32.9}{67.1}\times 6.40g$ of potassium.
So, potassium involved in reaction = 3.138 g
But we have given that the mass of potassium involved in the reaction is 3.60 g. Thus, potassium is excess reactant and bromine is limiting reactant.
Thus, moles of potassium which would actually combine with bromine to form KBr can be calculated as;
Weight of potassium reacting in the reaction = 3.138 g
Molecular weight of potassium = 39 g/mol
Moles of potassium reacted = $\dfrac{3.138g}{39g/mol}=0.08moles$
Therefore, 0.08 moles of K are reacted in the reaction with bromine to form KBr.

Note:
Do note that stoichiometry doesn’t play a role here as all the constituent masses and percentages are already given. But we need to consider the units of the calculating terms when we solve the illustration.