Chemistry Question

Question

If 75.0 of potassium hydroxide is permitted to react with 50.g of hydrochloric acid, how much potassium chloride is formed?

Answer 1

Hello!

First we need to start with a balanced equation:

Potassium hydroxide - KOH

Hydrochloric acid - HCl

This is a double displacement/acid base reaction

1 KOH + 1 HCl → 1 H₂O + 1 KCl

Now that we have the balanced equation, we need to convert the grams of each substance into moles of that substance and then into moles of KCl. This is called a Limiting Reactant problem, because you are given the mass of two of the reactants.

We know we have 75g of KOH, so let's convert that first into moles of KOH by using the molar mass found from the Periodic Table.

K = 39.1 g/mol, O = 16 g/mol, H = 1.01 g/mol, so the molar mass of KOH is all of them added together:

39.1 + 16 + 1.01 = 56.11 g/mol

To get from grams to moles, we will use some dimensional analysis/stoichiometry. If there's any question on whether to divide or multiply, divide or multiply your units and see if you get the units you want.

75 g KOH x 1 mole KOH = 1.34 mol KOH

1 56.11 g KOH

Now we can use our balanced equation to convert between moles of KOH and moles of KCl. The balanced equation gives us our "molar ratios" which tells you how many moles of the reactants are necessary to produce some number of moles of the product. So, because the leading coefficients are all 1 in the balanced equation, we can say that the amount of KOH is the exact amount of KCl that will be produced.

Therefore,

1.34 mol KOH = 1.34 mol KCl

Now let's do the same thing for the 50 g of HCl.

50 g HCl x 1 mole HCl = 1.37 mol HCl = 1.37 mol KCl

1 36.46 g HCl

So now we have two amounts of KCl. However, because this is a Limiting Reactant problem, we are only going to choose the smallest number because that's the amount that will be produced before the limiting reactant runs out.

Think of it like making a tuna melt. For every tuna melt you have two slices of bread, one can of tuna, and one slice of cheese. If you start with 50 pieces of bread, cans of tuna, and slices of cheese, the bread would run out first and the maximum amount of sandwiches you could make is 25 even though you still have cheese and tuna leftover. Same thing here, once the limiting reactant runs out, you cannot make anymore product.

So we will use the 1.34 moles of KCl because it's the smaller out of the two numbers, meaning that KOH would be the limiting reactant.

Therefore 1.34 moles of KCl are formed, or if we convert this into grams,

1.34 mole KCl x 74.55 g KCl = 99.9 g KCl

1 1 mole KCl