Q76P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the maximum numbers of moles and grams of ${\mathbf{H}}_{\mathbf{2}}\mathbf{S}$ that can form when 158 g of aluminium sulfide reacts with 131 g of water:

${\mathbf{AI}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{3}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\to }\mathbf{AI}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}\mathbf{\left[}\mathbf{unbalanced}\mathbf{\right]}$

What mass of the excess reactant remains?

Step-by-Step Solution

Answer

1. The number of moles of ${\mathbf{H}}_{\mathbf{2}}\mathbf{S}$  is 3.15mol.
2. The mass of ${\mathbf{H}}_{\mathbf{2}}\mathbf{S}$ is 107.415 g.
3. The mass of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ left is 17.48 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 stoichiometric coefficient.

So, the balanced chemical equation is:

${\mathbf{AI}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{3}}\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\to }\mathbf{2}\mathbf{AI}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}$

Step 2: 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.

$\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{moles}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{Molar}\mathbf{ }\mathbf{mass}}$

Step 3: Calculate the number of moles of AI 2 S 3

The mass of${}^{{\mathbf{AI}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{3}}}$  = 158 g.

The molar mass of${}^{{\mathbf{AI}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{3}}}$ = 150.158 g/mol

Thus, the number of moles ${}^{{\mathbf{AI}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{3}}}$  is:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{moles} {\mathrm{AI}}_2{\mathrm{S}}_3 = \frac{\mathrm{mass} \mathrm{of} {\mathrm{AI}}_2{\mathrm{S}}_3}{\mathrm{Molar} \mathrm{mass} \mathrm{of} {\mathrm{AI}}_2{\mathrm{S}}_3} \\ =\frac{158 \mathrm{g}}{150.158\mathrm{g}/\mathrm{mol}} \\ =1.05 \mathrm{mol} \end{gathered}$$

Step 4: Calculate the number of moles of H 2 0

The mass of ${}^{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}$  = 131 g.

The molar mass of ${}^{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}$  =18.02 g/mol

Thus, the number of moles ${}^{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}$ is:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{H}}_2\mathrm{O} = \frac{\mathrm{mass} \mathrm{of} {\mathrm{H}}_2\mathrm{O}}{\mathrm{Molar} \mathrm{mass} \mathrm{of} {\mathrm{H}}_2\mathrm{O}} \\ = \frac{131 \mathrm{g}}{18.02 \mathrm{g}/\mathrm{mol}} \end{gathered}$$

Step 5: Determine the Limiting reagent of the reaction

In the reaction, 1mol of ${}^{{\mathbf{AI}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{3}}}$ reacts with 6mol of ${}^{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}$ Thus,

  $$\begin{gathered} 1\mathrm{mol} \mathrm{of} {\mathrm{AI}}_2{\mathrm{S}}_3 = 3\mathrm{mol} \mathrm{of} {\mathrm{H}}_2\mathrm{S} \\ 1.05\mathrm{mol} \mathrm{of} {\mathrm{AI}}_2{\mathrm{S}}_3 = 1.05\times 3\mathrm{mol} \mathrm{of} {\mathrm{H}}_2\mathrm{S} \\ = 3.15\mathrm{mol} \mathrm{of} {\mathrm{H}}_2\mathrm{S} \end{gathered}$$

Therefore, ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ present in access in the reaction, hence ${}^{{\mathbf{AI}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{3}}}$ is a limiting reagent.

Step 6: Calculate the number of moles of H 2 S

In the given reaction,1mol of ${}^{{\mathbf{AI}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{3}}}$ forms 3mol of ${\mathbf{H}}_{\mathbf{2}}\mathbf{S}$ 

Thus, the number of moles of ${\mathbf{H}}_{\mathbf{2}}\mathbf{S}$ is

$$\begin{gathered} 1\mathrm{mol} \mathrm{of} {\mathrm{AI}}_2{\mathrm{S}}_3 = 3\mathrm{mol} \mathrm{of} {\mathrm{H}}_2\mathrm{S} \\ 1.05\mathrm{mol} \mathrm{of} {\mathrm{AI}}_2{\mathrm{S}}_3 = 1.05\times 3\mathrm{mol} \mathrm{of} {\mathrm{H}}_2\mathrm{S} \\ = 3.15\mathrm{mol} \mathrm{of} {\mathrm{H}}_2\mathrm{S} \end{gathered}$$

Step 7: Calculate the mass of H 2 S

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

Thus, the number of moles of${}^{{\mathbf{H}}_{\mathbf{2}}\mathbf{S}}$ is:

$$\begin{gathered} \mathrm{mass} \mathrm{of} {\mathrm{H}}_2\mathrm{S} = \mathrm{moles} \mathrm{of} {\mathrm{H}}_2\mathrm{S}\times \mathrm{molarmass} \mathrm{of} {\mathrm{H}}_2\mathrm{S} \\ = 3.15\mathrm{mol}\times 34.1\mathrm{g}/\mathrm{mol} \\ = 107.415 \mathrm{g} \end{gathered}$$

Step 8: Calculate the mass of H 2 O left

The molar mass of${}^{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}$ =18.02 g/mol

The number of moles${}^{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}$ left = 7.27mol – 6.3mol = 0.97mol

Thus, the mass of${}^{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}$ left is:

$$\begin{gathered} \mathrm{mass} \mathrm{of} {\mathrm{H}}_2\mathrm{Oleft} = \mathrm{moles} \mathrm{of} {\mathrm{H}}_2\mathrm{Oleft}\times \mathrm{molarmass} \mathrm{of} {\mathrm{H}}_2\mathrm{O} \\ = 0.97\mathrm{mol}\times 18.02\mathrm{g}/\mathrm{mol} \\ = 17.48 \mathrm{g} \end{gathered}$$