Q3.64P - Chemistry: Molecular Nature Of Matter And Change

Question

Bismuth oxide reacts with carbon to form bismuth metal: $\mathit{B}{\mathit{i}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}\left(s\right)\mathbf{+}\mathbf{3}\mathit{C}\left(s\right)\mathbf{\to }\mathbf{2}\mathit{B}\mathit{i}\left(s\right)\mathbf{+}\mathbf{3}\mathit{C}{\mathit{O}}_{\mathbf{2}}\left(g\right)$

When 283 g of $\mathit{B}{\mathit{i}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}$ reacts with excess carbon,

(a) How many moles of $\mathit{B}{\mathit{i}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}$ react?

(b) How many moles of $\mathit{B}\mathit{i}$ form?

Step-by-Step Solution

Answer

Answer

1. The number of moles of $\mathit{B}{\mathit{i}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}$ formed is 0.607 mol.

2. The number of moles of $\mathit{B}\mathit{i}$ formed is 1.214 mol.

Step 1: Relation between mass and number of moles

The number of moles is calculated by the mass and Molar mass. The relationship between the number of moles, mass, and molar mass is given below.

$\mathit{N}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{=}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{ }\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}$

Step 2: Conversion of mass of into moles

The mass of the $\mathit{B}{\mathit{i}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}$ = 283 g

The molar mass of $\mathit{B}{\mathit{i}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}$ = 465.96 g/mol

Thus, the number of moles of $\mathit{B}{\mathit{i}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}$ is:

$\overline{)\begin{array}{ccc}\mathbf{N}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{B}{\mathbf{i}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}& \mathbf{=}& \frac{\mathbf{283}\mathbf{ }\mathbf{g}}{\mathbf{465}\mathbf{.}\mathbf{96}\mathbf{ }\mathbf{g}\mathbf{/}\mathbf{m}\mathbf{o}\mathbf{l}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{60735}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{l}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{607}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{l}\end{array}}$

Step 3: Determine the Limiting reagent of the reaction

In the reaction, carbon is present in excess, so when 283 g of $\mathit{B}{\mathit{i}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}$ reacts with an excess of carbon, $\mathit{B}{\mathit{i}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}$ will be completely consumed in the reaction. Therefore, $\mathit{B}{\mathit{i}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}$ is a limiting reagent.

This states that the reactant which is completely utilized in the reaction or completely used up in the reaction acts as a Limiting reagent.

Step 4: Calculate the number of moles of Bi formed

In the given reaction, 1mol of $\mathit{B}{\mathit{i}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}$reacts with carbon and forms 2 mol of $\mathit{B}\mathit{i}\mathbf{.}$

Thus,

$\begin{array}{ccc}\mathbf{1}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{B}{\mathbf{i}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}& \mathbf{=}& \mathbf{2}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{B}\mathbf{i}\\ \mathbf{0}\mathbf{.}\mathbf{607}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{B}{\mathbf{i}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{607}\mathbf{\times }\mathbf{2}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{B}\mathbf{i}\\ & \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{214}\mathbf{ }\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{B}\mathbf{i}\end{array}$

Therefore, the number of moles of $\mathit{B}\mathit{i}$ formed is 1.214 mol.