Q3.124CP - Chemistry: Molecular Nature Of Matter And Change

Question

 The following circles represent a chemical reaction between AB2 and B2:

(a) Write a balanced equation for the reaction. (b) What is the limiting reactant? (c) How many moles of product can be made from 3.0 mol of B2 and 5.0 mol of AB2? (d) How many moles of excess reactant remain after the reaction in part (c)?

Step-by-Step Solution

Answer

Answer

(a) The balanced chemical reaction is 

(b) AB2 molecule is the limiting reactant.

(c) 5.0 moles of AB3 is formed.

(d) 0.5 moles of excess reactant are remained.

Step 1: (a) Writing the balanced chemical reaction

To write the balanced chemical reaction, first you have to count the number of molecules of the reactant and the product. Here, in the reactant there are 6 $A{B}_2$ and 5 $B_2$ molecules, and the product formed has 6 $A{B}_3$ and $2B_2$ molecules.

The reaction can be written as

$\mathbf{6}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{2}}\mathbf{+}\mathbf{5}\mathbf{ }{\mathit{B}}_{\mathbf{2}}\mathbf{ }\mathbf{\to }\mathbf{ }\mathbf{6}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{3}}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{2}{\mathit{B}}_{\mathbf{2}}$

By removing the extra B2, the equation can be written as

$\mathbf{6}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{2}}\mathbf{+}\mathbf{ }\mathbf{3}\mathbf{ }{\mathit{B}}_{\mathbf{2}}\mathbf{ }\mathbf{\to }\mathbf{ }\mathbf{6}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{3}}$

Diving by common coefficient, the final balanced equation is,

$\mathbf{2}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{2}}\mathbf{+}{\mathit{B}}_{\mathbf{2}}\mathbf{ }\mathbf{\to }\mathbf{ }\mathbf{2}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{3}}$

Step 2: (b) Description on limiting reactant

The reactant that limits the total amount of products that can form is called the limiting reactant. 

The equation shows that two $\mathit{A}{\mathit{B}}_{\mathbf{3}}$ molecules are formed for two $\mathit{A}{\mathit{B}}_{\mathbf{2}}$ molecule and single ${\mathit{B}}_{\mathbf{2}}$ molecules that react. To find the limiting reactant, you need to compare the number of molecules of each reactant, with number you need to react completely. Before the reaction, there were two ${\mathit{B}}_{\mathbf{2}}$ molecules remaining, so ${\mathit{B}}_{\mathbf{2}}$ is in excess. All the $\mathit{A}{\mathit{B}}_{\mathbf{2}}$ molecules have reacted so AB2 molecule is the limiting reactant.

Step 3: (c) Finding number of moles of A B 3

Finding moles of AB3 from the moles of $A{B}_2$ (if $B_2$ is limiting):  

$$\begin{gathered} \mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{3}}\mathbf{ }\mathit{f}\mathit{r}\mathit{o}\mathit{m}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{2}}\mathbf{=}\left(5.0 mol of A{B}_2\right)\left(\frac{2 mol A{B}_3}{2 mol A{B}_2}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{3}} \end{gathered}$$

Finding moles of AB3 from the moles of B2 (if AB2 is limiting):

$$\begin{gathered} \mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{3}}\mathbf{ }\mathit{f}\mathit{r}\mathit{o}\mathit{m}\mathbf{ }{\mathit{B}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{B}}_{\mathbf{2}}\mathbf{)}\mathbf{\left(}\frac{\mathbf{2}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{AB}}_{\mathbf{3}}}{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{B}}_{\mathbf{2}}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{3}} \end{gathered}$$

Whichever reactant forms the fewer moles of $A{B}_3$ is the limiting reactant.

Reactant molar ratio in the balanced chemical reaction determines which reactant is limiting.

AB2 is the limiting reactant and 5.0 mol of $A{B}_3$ is formed.

Step 3: (d) Finding number of moles of A B 3

To find the moles of excess reactant that remains, you will calculate the difference between the initial moles of excess reactant and the moles required for the reaction.

$$\begin{gathered} \mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }{\mathit{B}}_{\mathbf{2}}\mathbf{ }\mathit{r}\mathit{e}\mathit{a}\mathit{c}\mathit{t}\mathit{s}\mathbf{ }\mathit{w}\mathit{i}\mathit{t}\mathit{h}\mathbf{ }\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{2}}\mathbf{=}\left(5.0 mol of A{B}_2\right)\left(\frac{1 mol B_2}{2 mol A{B}_2}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathbf{ }\mathit{A}{\mathit{B}}_{\mathbf{3}} \end{gathered}$$

The excess reactant, B2 remained is,

                                3.0 mol – 2.5 mol       = 0.5 moles of B2