If 15 grams of barium chloride reacts with 20.0 grams of potassium sulfate, how much barium...

Given Data:

• The mass of barium chloride is 15 g.
• The mass of potassium sulfate is 20.0 g.

The balanced reaction is given below.

{eq}{\rm{BaC}}{{\rm{l}}_{\rm{2}}}\left( {{\rm{aq}}} \right) + {{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\left( {{\rm{aq}}} \right) \to {\rm{BaS}}{{\rm{O}}_{\rm{4}}}\left( {\rm{s}} \right) + {\rm{2KCl}}\left( {{\rm{aq}}} \right) {/eq}

The molar mass of barium chloride is {eq}208.23\;{\rm{g/mol}} {/eq}.

The moles of barium chloride can be calculated by the formula given below.

{eq}{\rm{Moles}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar}}\;{\rm{mass}}}}......\left( {\rm{I}} \right) {/eq}

Substitute the respective values in equation (I).

{eq}\begin{align*} {\rm{Moles}} &= \dfrac{{15\;{\rm{g}}}}{{208.23\;{\rm{g/mol}}}}\\ &= 0.07204\;{\rm{mol}} \end{align*} {/eq}

The moles of barium chloride are {eq}0.07204\;{\rm{mol}} {/eq}.

The molar mass of potassium sulfate is {eq}174.26\;{\rm{g/mol}} {/eq}.

The moles of potassium sulfate can be calculated by substituting the respective values in equation (I).

{eq}\begin{align*} {\rm{Moles}} &= \dfrac{{20.0\;{\rm{g}}}}{{174.26\;{\rm{g/mol}}}}\\ &= 0.1147\;{\rm{mol}}\\ &\approx {\rm{0}}{\rm{.115}}\;{\rm{mol}} \end{align*} {/eq}

The moles of potassium sulfate are {eq}{\rm{0}}{\rm{.115}}\;{\rm{mol}} {/eq}.

The limiting reagent of this can be evaluated by the division of calculated moles of barium chloride and potassium sulfate by their stoichiometric coefficients.

The stoichiometric coefficient of barium chloride is 1, so the moles of barium chloride are shown below.

{eq}\begin{align*} {\rm{Moles of barium}}\;{\rm{chloride}} &= \dfrac{{0.07204\;{\rm{mol}}}}{1}\\ &= 0.07204\;{\rm{mol}} \end{align*} {/eq}

The stoichiometric coefficient of potassium sulfate is 1, so the moles of potassium sulfate are shown below.

{eq}\begin{align*} {\rm{Moles of potassium}}\;{\rm{sulfate}} &= \dfrac{{{\rm{0}}{\rm{.115}}\;{\rm{mol}}}}{1}\\ &= {\rm{0}}{\rm{.115}}\;{\rm{mol}} \end{align*} {/eq}

Hence, the number of moles of barium chloride is less. So, the limiting reagent in this reaction is barium chloride.

From the above reaction, 1 mole of barium chloride gives 1 mole of barium sulfate.

So, the moles of barium sulfate is {eq}1 \times 0.07204\;{\rm{mol}} = 0.07204\;{\rm{mol}} {/eq}.

The moles of barium sulfate are {eq}0.07204\;{\rm{mol}} {/eq}.

The molar mass of barium sulfate is {eq}233.38\;{\rm{g/mol}} {/eq}.

The mass of barium sulfate can be calculated by the formula given below.

{eq}{\rm{Mass}} = {\rm{Moles}} \times {\rm{Molar}}\;{\rm{mass}}......\left( {{\rm{II}}} \right) {/eq}

Substitute the respective values in equation (II).

{eq}\begin{align*} {\rm{Mass}} &= 0.07204\;{\rm{mol}} \times 233.38\;{\rm{g/mol}}\\ &= 16.8126\;{\rm{g}}\\ &\approx {\rm{16}}{\rm{.813}}\;{\rm{g}} \end{align*} {/eq}

Therefore, the mass of barium sulfate is {eq}{\rm{16}}{\rm{.813}}\;{\rm{g}} {/eq}.