# Question 45871

May 20, 2017

${\text{0.5 g BaSO}}_{4}$

#### Explanation:

Start by writing a balanced chemical equation for your reaction

$\textcolor{b l u e}{3} {\text{BaCl"_ (2(aq)) + "Al"_ 2("SO"_ 4)_ (3(aq)) -> 3"BaSO"_ (4(s)) darr + 2"AlCl}}_{3 \left(a q\right)}$

You know that the two reactants take part in the reaction in a $\textcolor{b l u e}{3} : 1$ mole ratio, i.e. the reaction consumes $\textcolor{b l u e}{3}$ moles of barium chloride for every $1$ mole of aluminium sulfate that takes part in the reaction.

Use the molarities and the volumes of the two solutions to calculate the number of moles of each reactant available.

20 color(red)(cancel(color(black)("mL"))) * "0.1 moles BaCl"_2/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0020 moles BaCl"_2

30 color(red)(cancel(color(black)("mL"))) * ("0.2 moles Al"_ 2("SO"_ 4)_ 3)/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0060 moles Al"_2("SO"_4)_3

Now, pick a reactant and use the aforementioned mole ratio to see if you have enough moles of the second reactant to ensure that all the moles of the first reactant can take part in the reaction

0.0020 color(red)(cancel(color(black)("moles BaCl"_2))) * ("1 mole Al"_2("SO"_4)_3)/(color(blue)(3)color(red)(cancel(color(black)("moles BaCl"_2)))) = "0.00067 moles Al"_2("SO"_4)_3

As you can see, you have much more aluminium sulfate available than needed

$\text{0.0060 moles available " > " 0.00067 moles needed}$

which implies that aluminium sulfate is in excess. In other words, barium sulfate will be the limiting reagent, i.e. it will be completely consumed by the reaction before all the moles of aluminium sulfate will get the chance to react.

Notice that you have a $\textcolor{b l u e}{3} : 3$ mole ratio between barium chloride and barium sulfate, the precipitate.

This means that the reaction will produce

0.0020 color(red)(cancel(color(black)("moles BaCl"_2))) * "3 moles BaSO"_4/(color(blue)(3)color(red)(cancel(color(black)("moles BaCl"_2)))) = "0.0020 moles BaSO"_4#

Finally, to convert this to grams, use the molar mass of barium sulfate

$0.0020 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles BaSO"_4))) * "233.38 g"/(1color(red)(cancel(color(black)("mole BaSO"_4)))) = color(darkgreen)(ul(color(black)("0.5 g BaSO}}_{4}}}}$

The answer is rounded to one significant figure.