For the following reaction, 27.5 grams of hydrochloric acid are allowed to react with 56.7 grams of barium hydroxide.
hydrochloric acid (aq) + barium hydroxide (aq) > barium chloride (aq) + water (l)
(i) What is the maximum amount of barium chloride that can be formed in grams?
(ii) What is the FORMULA for the limiting reagent?
(iii) What amount of the excess reagent remains after the reaction is complete in grams?

i) 68.92 g

ii) Ba(OH)₂

iii) 3.35 g

Explanation:

This is a chemical reaction where an acid and a base are reacting. To get the mass of the product, in this case, the barium chloride, we need to write the reaction with the correct formula of the compound and then, balance, if it's neccesary.

The chemical reaction with formulas is:

HCl + Ba(OH)₂ > BaCl₂ + H₂O

Now, it's time to balance the equation:

2HCl + Ba(OH)₂ > BaCl₂ + 2H₂O

We have the balanced equation, now let's find out how much of the product will be formed. In this case, we have to use stochiometry with moles, which is the easier way to get the quantity of the products. With the balanced equation we can know which is the limiting reactant and the excess. Let's get the moles of the acid and the hydroxide.

The molecular mass of HCl and hydroxide reported is 36.45 g/mol and 171.34 g/mol so the moles are:

n HCl = 27.5/36.45 = 0.754 moles

n Ba(OH)₂ = 56.7/171.34 = 0.331 moles

Now, let's find the stochiometry to get the limiting reactant:

2 mole HCl > 1 mole Ba(OH)₂

0.754 moles > X

X = 0.754 / 2 = 0.377 moles of Ba(OH)₂ are needed, but we only have 0.331 moles, this means that the Ba(OH)₂ is the limiting reactant while the HCl is the excess reactant.

Now that we know that the excess reagent is the HCl, let's see how much of it remains after the reaction is completed:

moles of HCl that reacted: 0.331 * 2 = 0.662 moles

remanent moles of HCl = 0.754 - (0.662) = 0.092 moles

Then the mass:

m = 0.092 * 36.45 = 3.35 g of HCl

Now, let's see how much of BaCl₂ is formed, knowing that the moles of Ba(OH)₂ are the same moles of BaCl₂:

moles Ba(OH)₂ = moles BaCl₂ = 0.331 moles

The reported molar mass of BaCl₂ is 208.23 g/mol so the mass:

m BaCl₂ = 0.331 * 208.23 = 68.92 g

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