Question

How many millilitres of $0.125{\text{M}} {\text{Ba}}{({\text{OH}})_2}({\text{aq}})$ must be used to produce $5.5$ moles of water? Please balance the equation before solving the problem.

Answer 1

First write the chemical balanced equation. Then to calculate the moles of ${\text{Ba}}{({\text{OH}})_2}$ , use the molar ratio of the balanced equation. After that, we will calculate the volume of ${\text{Ba}}{({\text{OH}})_2}$ from the molarity given.

Formula Used: ${\text{molarity}} = \dfrac{{{\text{moles}}}}{{{\text{volume of solution}}}}$

Complete Step-by-Step solution
According to the question, the molarity of ${\text{Ba}}{({\text{OH}})_2}$ is given as $0.125 {\text{M}}$
Let us consider that $HCl$ is being used an acid here in this case
Then the balanced chemical reaction would be
${\text{Ba}}{({\text{OH}})_2} + 2{\text{HCl}} \to {\text{BaC}}{{\text{l}}_2} + 2{{\text{H}}_2}{\text{O}}$
Now, we will proceed to calculate the number of moles of ${\text{Ba}}{({\text{OH}})_2}$
The molar ratio from the balanced equation is
${\text{Ba}}{({\text{OH}})_2}:{{\text{H}}_2}{\text{O}} = 1:2$
We can easily calculate the number of moles of ${\text{Ba}}{({\text{OH}})_2}$ by simple unitary method as the stoichiometric coefficients are already taken out by us
Moles of ${\text{Ba}}{({\text{OH}})_2} = 5.5 {\text{mol}} {{\text{H}}_2}{\text{O}} \times \dfrac{{1{\text{mol}} {\text{Ba}}{{({\text{OH}})}_2}}}{{2{\text{mol}} {{\text{H}}_2}{\text{O}}}} = 2.75 {\text{molBa}}{({\text{OH}})_2}$
the molarity of ${\text{Ba}}{({\text{OH}})_2}$ is given as $0.125 {\text{M}}$ , this means that volume of $0.125{\text{M}} {\text{Ba}}{({\text{OH}})_2}({\text{aq}}) = 1 L$
Now, in the same way as the previous, we can calculate the required volume
$0.125{\text{M}} {\text{Ba}}{({\text{OH}})_2}({\text{aq}})$ contains $1 L$ of the solution
$2.75 {\text{mol Ba}}{({\text{OH}})_2}$ will contain:
${\text{volume of Ba}}{({\text{OH}})_2} = \dfrac{{2.75{\text{mol Ba}}{{({\text{OH}})}_2}}}{{0.125{\text{M Ba}}{{({\text{OH}})}_2}}}$
Upon solving, we get Volume of ${\text{Ba}}{({\text{OH}})_2} = 22{\text{L}} {\text{Ba}}{({\text{OH}})_2}$
Now, we will convert this to millilitres
Hence, the required volume in millilitres is $22000{\text{ mL Ba(OH}}{{\text{)}}_{\text{2}}}$ .

Additional Information:
The proportions of reactants and products that are used and formed in a chemical reaction are stated by molar ratios. It is possible to derive molar ratios from the coefficients of a balanced chemical equation.

Note:
In solutions, the majority of reactions occur and it is therefore essential to understand how the amount of substance is expressed when it is present in the solution. The quantity of substances in the solution is expressed in many ways.