Question

Calculate the mass of each product formed when 174 g of silver sulphide reacts with excess hydrochloric acid:

$\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}\left(s\right)\mathbf{+}\mathit{H}\mathit{C}\mathit{l}\left(aq\right)\mathbf{\to }\mathit{A}\mathit{g}\mathit{C}\mathit{l}\left(s\right)\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{S}\left(g\right)\left[unbalanced\right]$

Short answer

Answer

1. The mass of $\mathit{A}\mathit{g}\mathit{C}\mathit{l}$is 201.28 g.

2. The mass of ${\mathit{H}}_{\mathbf{2}}\mathit{S}$is 23.9 g.

Step-by-step solution

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 stoichiometric coefficient.

So, the balanced chemical equation is:

$\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}\left(s\right)\mathbf{+}\mathbf{2}\mathit{H}\mathit{C}\mathit{l}\left(aq\right)\mathbf{\to }\mathbf{2}\mathit{A}\mathit{g}\mathit{C}\mathit{l}\left(s\right)\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{S}\left(g\right)$

Determine the Limiting reagent of the reaction

In the reaction, $\mathit{H}\mathit{C}\mathit{l}$is present in excess, so when 174 g of $\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}$reacts with an excess of $\mathit{H}\mathit{C}\mathit{l}\mathbf{,}\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}$will be completely consumed in the reaction. Therefore, $\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}$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.

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.

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

Calculate the number of moles of Ag2S formed

The mass of $\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}$= 174 g.

The molar mass of $\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}$= 247.8 g/mol.

The number of moles of $\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}$is calculated as:

$$\begin{gathered} \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}\mathbf{=}\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{Ag}}_{\mathbf{2}}\mathbf{S}}{\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{Ag}}_{\mathbf{2}}\mathbf{S}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{174}\mathbf{}\mathbf{g}}{\mathbf{247}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7022}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

Therefore, the number of moles of $\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}$is 0.7022 mol.

Relation between the number of moles of Ag2Sand AgCl

In the given reaction, 2mol of $\mathit{A}\mathit{g}\mathit{C}\mathit{l}$is formed from 1mol of$\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}\mathbf{.}$

Thus,

$\mathbf{1}\mathit{m}\mathit{o}\mathit{l}\mathit{o}\mathit{f}\mathbf{}\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}\mathbf{=}\mathbf{2}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{A}\mathit{g}\mathit{C}\mathit{l}$

Calculate the number of moles of AgCl

The number of moles of $\mathit{A}\mathit{g}\mathit{C}\mathit{l}$is:

$$\begin{gathered} 1molofA{g}_{2}S=2molofAgCl \\ 0.7022molofA{g}_{2}S=0.7022\times 2molofAgCl \\ =14044molofAgCl \end{gathered}$$

Calculate the mass of AgCl

The molar mass of $\mathit{A}\mathit{g}\mathit{C}\mathit{l}$= 143.32 g/mol.

The mass of $\mathit{A}\mathit{g}\mathit{C}\mathit{l}$is calculated as:

$$\begin{gathered} \mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{A}\mathit{g}\mathit{C}\mathit{l}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{A}\mathit{g}\mathit{C}\mathit{l}\mathbf{\times }\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{}\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{A}\mathit{g}\mathit{C}\mathit{l} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{4044}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathbf{\times }\mathbf{143}\mathbf{.}\mathbf{32}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{201}\mathbf{.}\mathbf{28}\mathbf{}\mathit{g} \end{gathered}$$

Therefore, the mass of $\mathit{A}\mathit{g}\mathit{C}\mathit{l}$is approximately 201.28 g.

Relation between the number of moles of Ag2S and H2S

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

Thus,

$\mathbf{1}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{A}{\mathit{g}}_{\mathbf{2}}\mathit{S}\mathbf{=}\mathbf{1}\mathit{m}\mathit{o}\mathit{l}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathit{H}}_{\mathbf{2}}\mathit{S}$

Calculate the number of moles of H2S

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

$\overline{)\begin{array}{c}1molofA{g}_{2}S=1molof{H}_{2}S\\ 0.7022molofA{g}_{2}S=0.702molof{H}_{2}S\end{array}}$

Calculate the mass of H2S

The molar mass of ${\mathit{H}}_{\mathbf{2}}\mathit{S}$= 34.1 g/mol.

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

$$\begin{gathered} \mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathit{H}}_{\mathbf{2}}\mathit{S}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathit{H}}_{\mathbf{2}}\mathit{S}\mathbf{\times }\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{}\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathit{H}}_{\mathbf{2}}\mathit{S} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7022}\mathit{m}\mathit{o}\mathit{l}\mathbf{\times }\mathbf{34}\mathbf{.}\mathbf{1}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{23}\mathbf{.}\mathbf{945}\mathit{g} \end{gathered}$$

Therefore, the mass of ${\mathit{H}}_{\mathbf{2}}\mathit{S}$is approximately 23.9 g.