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Determine the mass (in g) of NiCO3 that is produced when 770 mL of a 4.71 x 10-2 M Na2CO3...

Question:

Determine the mass (in g) of {eq}NiCO_3 {/eq} that is produced when 770 mL of a {eq}4.71 \times 10^{-2} {/eq} M {eq}Na_2CO_3 {/eq} solution completely reacts with 626 mL of a {eq}6.22 \times 10^{-2} {/eq} M {eq}NiCl_2 {/eq} solution according to the following balanced chemical equation.

{eq}Na_2CO_3(aq) + NiCl_2(aq) \to NiCO_3(s) + 2NaCl(aq) {/eq}

Limiting Reagent:

For a given reaction, out of the many reactants, the product amount depends upon only the amount of the limiting reagent. So, it necessary to identify the limiting reagent by determining the reactant that is present in the least quantity to predict the product amount.

Answer and Explanation: 1

Given Data:

  • The volume of sodium carbonate is 770 mL.
  • The molarity of sodium carbonate is {eq}4.71 \times {10^{ - 2}} {/eq} M.
  • The volume of nickel carbonate is 626 mL.
  • The molarity of nickel carbonate is {eq}6.22 \times {10^{ - 2}} {/eq} M.


The given reaction is shown below:

{eq}{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_3}\left( {{\rm{aq}}} \right) + {\rm{NiC}}{{\rm{l}}_2}\left( {{\rm{aq}}} \right) \to {\rm{NiC}}{{\rm{O}}_3}\left( {\rm{s}} \right) + 2\,{\rm{NaCl}}\left( {{\rm{aq}}} \right) {/eq}


Convert the given volumes from mL to L.

Sodium carbonate:

{eq}\begin{align*} {\rm{1000}}\,{\rm{mL}} &= 1\,{\rm{L}}\,\\ {\rm{770}}\,{\rm{mL}} &= 0.770\,\,{\rm{L}} \end{align*} {/eq}

Nickel chloride:

{eq}\begin{align*} {\rm{1000}}\,{\rm{mL}} &= 1\,{\rm{L}}\,\\ 626\,\,{\rm{mL}} &= 0.626\,{\rm{L}} \end{align*} {/eq}


Calculate the moles of sodium carboante and nickel chloride using the molarity formula as follows:

{eq}{\rm{Molarity}} = \dfrac{{{\rm{Moles}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{solution}}}} {/eq}


Sodium carbonate:

Substitute the known values.

{eq}\begin{align*} 4.71 \times {10^{ - 2}}\,{\rm{M}} &= \dfrac{{{\rm{Mole}}}}{{0.770\,{\rm{L}}}}\\ {\rm{Mole}} &= 4.71 \times {10^{ - 2}}\,{\rm{M}} \times 0.770\,{\rm{L}}\\ &= 0.036\,{\rm{mol}} \end{align*} {/eq}


Nickel chloride:

Substitute the known values.

{eq}\begin{align*} 6.22 \times {10^{ - 2}}\,{\rm{M}} &= \dfrac{{{\rm{Mole}}}}{{0.626\,{\rm{L}}}}\\ {\rm{Mole}} &= 6.22 \times {10^{ - 2}}\,{\rm{M}} \times 0.626\,{\rm{L}}\\ &= 0.038\,{\rm{mol}} \end{align*} {/eq}


Limiting reagent is depicted as follows:

From above given reaction, we can say 1 mole sodium carbonate reacts with 1 mole of nickel chloride. So, 0.036 moles of sodium carbonate will react with;

{eq}\begin{align*} {\rm{1}}\,{\rm{mol}}\,{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_3} &= 1\,{\rm{mol}}\,{\rm{NiC}}{{\rm{l}}_2}\,\\ 0.036\,{\rm{mol}}\,{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_3} &= 0.036\,\,{\rm{mol}}\,{\rm{NiC}}{{\rm{l}}_2} \end{align*} {/eq}

As the sodium carbonate is present in less amount so, sodium carbonate is the limiting reagent.


The amount of product (nickel carbonate) is depicted as follows:

From above given reaction, we can say 1 mole sodium carbonate gives 1 mole of nickel carbonate. So, 0.036 moles of sodium carbonate will give;

{eq}\begin{align*} {\rm{1}}\,{\rm{mol}}\,{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_3} &= 1\,{\rm{mol}}\,{\rm{NiC}}{{\rm{O}}_3}\\ 0.036\,{\rm{mol}}\,{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_3} &= 0.036\,\,{\rm{mol}}\,{\rm{NiC}}{{\rm{O}}_3} \end{align*} {/eq}


Calculate the mass of nickel carboante using the mole formula as follows:

{eq}{\rm{Mole}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar}}\,{\rm{mass}}}} {/eq}

The molar mass of nickel carbonate is 118.70 g/mol.

Substitute the known values.

{eq}\begin{align*} {\rm{0}}{\rm{.036}}\,{\rm{mol}} &= \dfrac{{{\rm{Mass}}}}{{118.70\,{\rm{g/mol}}}}\\ {\rm{Mass}} &= {\rm{0}}{\rm{.036}}\,{\rm{mol}} \times 118.70\,{\rm{g/mol}}\\ &= 4.27\,{\rm{g}} \end{align*} {/eq}


Therefore, 4.27 g nickel carbonate will be produced.


Learn more about this topic:

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Limiting Reactants & Calculating Excess Reactants

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Chapter 9 / Lesson 5
31K

Learn how to find the limiting and excess reactants in a chemical reaction. See example problems that calculate the limiting and excess reactants.


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