Q3.67P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the mass of each product formed when 43.82 g of diborane$\left(B_2{H}_6\right)$reacts with excess water:

${\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{I}\mathbf{\right)}\mathbf{\to }{\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\left[}\mathbf{unbalanced}\mathbf{\right]}$

Step-by-Step Solution

Answer

Answer

1. The mass of ${\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}$ is 195.4 g.

2. The mass of ${\mathit{H}}_{\mathbf{2}}$ is 9.55 g.

Step 1: Balance the chemical equation

In a balanced chemical equation, the number of atoms on the reactant side should be the same as the number of atoms on the product side for a particular atom. For balancing the reaction, the atoms are multiplied by the stochiometric coefficient.

So, the balanced chemical equation is:

${\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\left(g\right)\mathbf{+}\mathbf{6}{\mathit{H}}_{\mathbf{2}}\mathit{O}\left(I\right)\mathbf{\to }\mathbf{2}{\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}\left(s\right)\mathbf{+}\mathbf{6}{\mathit{H}}_{\mathbf{2}}\left(g\right)$

Step 2: Determine the Limiting reagent of the reaction

In the reaction, water is present in excess, so when 43.82 g of ${\mathit{H}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}$ reacts with an excess of water, ${\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}$ will be completely consumed in the reaction. Therefore, ${\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}$ 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 3: 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 4: Calculate the number of moles of B 2 H 6 formed

The mass of ${\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}$= 43.82 g.

The molar mass of ${\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}$ = 27.66 g/mol.

The number of moles of ${\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}$is calculated as:

$$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }{\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{=}\frac{\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{B}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}}{\mathbf{Molar}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{B}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}} \\ \mathbf{=}\frac{\mathbf{43}\mathbf{.}\mathbf{82}\mathbf{g}}{\mathbf{27}\mathbf{.}\mathbf{66}\mathbf{ }\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{58}\mathbf{ }\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

Therefore, the number of moles of ${\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}$ formed is 1.58 mol.

Step 5: Relation between the number of moles of B 2 H 6 and H 3 B O 3

In the given reaction, 2 mol of ${\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}$ is formed from 1 mol of ${\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{.}$

Thus,

$\mathbf{1}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}{\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{=}\mathbf{2}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}{\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}$

Step 6: Calculate the number of moles of H 3 B O 3

The number of moles of ${\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}$is:

$$\begin{gathered} \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{1}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}{\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{2}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}{\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}} \\ \mathbf{1}\mathbf{.}\mathbf{58}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}{\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{58}\mathbf{\times }\mathbf{2}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}{\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}} \\ \mathbf{=}\mathbf{3}\mathbf{.}\mathbf{16}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}{\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}} \end{gathered}$$

Step 7: Calculate the mass of H 3 B O 3

The molar mass of ${\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}$= 61.83 g/mol.

The mass of ${\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}$ is calculated as:

$$\begin{gathered} \mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }{\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}\mathbf{=}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{o}\mathit{f}{\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}\mathbf{\times }\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{ }\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }{\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}} \\ \mathbf{=}\mathbf{3}\mathbf{.}\mathbf{16}\mathit{m}\mathit{o}\mathit{l}\mathbf{\times }\mathbf{61}\mathbf{.}\mathbf{83}\mathbf{ }\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \\ \mathbf{=}\mathbf{195}\mathbf{.}\mathbf{3838}\mathit{g} \end{gathered}$$

Therefore, the mass of ${\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}$is approximately195.4 g.

Step 8: Relation between the number of moles of B 2 H 6 and H 2

In the given reaction, 6 mol of ${\mathit{H}}_{\mathbf{2}}$ is formed from 1 mol of ${\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}$.

Thus,

$\mathbf{1}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}{\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{=}\mathbf{6}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}{\mathit{H}}_{\mathbf{2}}$

Step 9: Calculate the number of moles of H 2

The number of moles of ${\mathit{H}}_{\mathbf{2}}$is:

$$\begin{gathered} \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{1}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}{\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{6}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}\mathbf{ }{\mathit{H}}_{\mathbf{2}} \\ \mathbf{1}\mathbf{.}\mathbf{58}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }{\mathit{B}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{58}\mathbf{\times }\mathbf{6}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}\mathbf{ }{\mathit{H}}_{\mathbf{2}} \\ \mathbf{=}\mathbf{9}\mathbf{.}\mathbf{48}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}{\mathit{H}}_{\mathbf{2}} \end{gathered}$$

Step 10: Calculate the mass of H 2

The molar mass of ${\mathit{H}}_{\mathbf{2}}$= 1.00784 g/mol.

The mass of ${\mathit{H}}_{\mathbf{2}}$ is calculated as:

$$\begin{gathered} \mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }{\mathit{H}}_{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }{\mathit{H}}_{\mathbf{2}}\mathbf{\times }\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{ }\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }{\mathit{H}}_{\mathbf{2}} \\ \mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{9}\mathbf{.}\mathbf{48}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{00784}\mathbf{ }\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \\ \mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{9}\mathbf{.}\mathbf{5543232}\mathit{g} \end{gathered}$$

Therefore, the mass of ${\mathit{H}}_{\mathbf{2}}$ is approximately9.55 g.