For the following reaction, 30.3 grams of hydrochloric acid are allowed to react with 61.0 grams of barium hydroxide .

Question

hydrochloric acid(aq) + barium hydroxide(aq) barium chloride(aq) + water(l)

What is the maximum amount of barium chloride that can be formed? grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

First, write the balanced reaction equation:  2HCl(aq) + Ba(OH)2(aq) ==> BaCl₂(aq) + 2H₂O(l)

Next, find the limiting reactant.  This is most easily done by comparing moles HCl to moles Ba(OH)₂ and dividing each by the coefficient in the balanced equation.

>> Moles HCl = 30.3 g x 1 mole/36.5 g = 0.830 moles HCl (divided by 2 = 0.415 moles)

>> Moles Ba(OH)₂ = 61.0 g x 1 mole/171g = 0.357 moles Ba(OH)₂ (divided by 1 = 0.357 moles)

Since 0.357 is less than 0.415, Ba(OH)₂ is limiting.

Maximum amount of BaCl₂ that can be formed (in grams) is determined by the amount of Ba(OH)₂

0.357 moles Ba(OH)₂ x 1 mole BaCl₂/mole Ba(OH)₂ = 0.357 moles BaCl₂ can be formed

Mass = 0.357 moles BaCl₂ x 208 g/mole = 74.2 grams of BaCl₂ can be formed

 

Formula of limiting reagent is Ba(OH)₂

 

Amount of excess reagent (HCl) is determined as follows:

0.357 moles Ba(OH)₂ x 2 moles HCl/mole Ba(OH)₂ = 0.714 moles HCl has reacted.

moles HCl left over = 0.830 moles - 0.714 moles = 0.116 moles HCl left over

grams HCl left over = 0.116 moles HCl x 36.5 g/mole = 4.23 grams HCl left over.