Q50P - Chemistry: Molecular Nature Of Matter And Change

Question

The scenes below represent a chemical reaction between elements A (red) and B (green):

Which best represents the balanced equation for the reaction?

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{ }\mathbf{2}\mathbf{A}\mathbf{\to }{\mathbf{A}}_{\mathbf{2}}\mathbf{+}{\mathbf{B}}_{\mathbf{2}} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{ }{\mathbf{A}}_{\mathbf{2}}\mathbf{+}{\mathbf{B}}_{\mathbf{2}}\mathbf{\to }\mathbf{2}\mathbf{AB} \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{ }{\mathbf{B}}_{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{AB}\mathbf{\to }\mathbf{2}{\mathbf{B}}_{\mathbf{2}}\mathbf{+}{\mathbf{A}}_{\mathbf{2}} \\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{ }\mathbf{4}{\mathbf{A}}_{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{B}}_{\mathbf{2}}\mathbf{\to }\mathbf{8}\mathbf{AB} \end{gathered}$$

Step-by-Step Solution

Answer

The circle containing a mixture of  _{$A_2$}and $B^2$before reacting to form AB

In this case (d) will be the correct representation for the reaction because there are 4 moles of $A_2$ and as well as $B_2$ are present which further combined to form 8 moles of product, this can be observed in the diagram represented.

Mixed colour shows the formation of product by combining of both $A_2$ and and $B^2$forming AB as product.

Step 1: Finding the number of moles

In the given figure, two red color and two green color elements are present in the reactant sides. This indicates that divalent molecules are present in the reactant side. Since, 4 green color divalent molecule and 4 red color divalent molecule combine and form product. In the product side, two element combine and show both red and green color compound.

Step 2: Determine the balanced chemical equation for the reaction

In the given representation, 4 moles of reacts with 4 moles of $B_2$ and form 8 moles of AB product.

Therefore, the balanced chemical equation for the reaction is:

$\mathbf{4}{\mathbf{A}}_{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{B}}_{\mathbf{2}}\mathbf{\to }\mathbf{8}\mathbf{AB}$

Hence, option (d) is correct.