Q74P - Chemistry: Molecular Nature Of Matter And Change

Question

Metal hydrides react with water to form hydrogen gas and the metal hydroxide. For example, ${\mathbf{SrH}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{I}\mathbf{\right)}\mathbf{\to }\mathbf{Sr}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$ 

You wish to calculate the mass of hydrogen gas that can be prepared from 5.70 g of ${\mathbf{SrH}}_{\mathbf{2}}$ and 4.75 g of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ 

(a) How many moles of ${\mathbf{H}}_{\mathbf{2}}$ can be produced from the given mass of ${\mathbf{SrH}}_{\mathbf{2}}$

(b) How many moles of ${\mathbf{H}}_{\mathbf{2}}$ can be produced from the given mass of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ 

(c) Which is the limiting reactant? 

(d) How many grams of  ${\mathbf{H}}_{\mathbf{2}}$ can be produced?

Step-by-Step Solution

Answer

(a) The number of moles of ${\mathbf{H}}_{\mathbf{2}}$ produced from is 0.1272 mol.

(b) The number of moles of ${\mathbf{H}}_{\mathbf{2}}$ produced from is 0.0636 mol.

(c) $\mathbf{Sr}{\mathbf{H}}_{\mathbf{2}}$ is the limiting reactant.

(d) The mass of ${\mathbf{H}}_{\mathbf{2}}$ produced is 0.1282 g.

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.

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

Step 2: Calculate the number of moles of SrH 2

The mass of ${\mathbf{SrH}}_{\mathbf{2}}$  = 5.70 g.

The molar mass of ${\mathbf{SrH}}_{\mathbf{2}}$ = 89.64 g/mol

Thus, the number of moles is ${\mathbf{SrH}}_{\mathbf{2}}$:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{moles}=\frac{\mathrm{mass} {\mathrm{SrH}}_2}{\mathrm{Molar} \mathrm{mass} {\mathrm{SrH}}_2} \\ =\frac{5.70 \mathrm{g}}{89.64 \mathrm{g}/\mathrm{mol}} \\ =0.0636\mathrm{mol} \end{gathered}$$

Step 3: Calculate the number of moles of H 2 O

The mass of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$  = 4.75 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:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{moles} {\mathrm{H}}_2\mathrm{O}=\frac{\mathrm{mass} {\mathrm{H}}_2\mathrm{O}}{\mathrm{Molar} \mathrm{mass} {\mathrm{H}}_2\mathrm{O}} \\ =\frac{4.75 \mathrm{g}}{18.02 \mathrm{g}/\mathrm{mol}} \\ =0.2636\mathrm{mol} \end{gathered}$$

Step 4: Determine the Limiting reagent of the reaction

In the reaction, 1 mol of ${\mathbf{SrH}}_{\mathbf{2}}$  reacts with 2 mol of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ Thus,

$$\begin{gathered} \mathbf{1}\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{SrH}}_{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{2}\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ \mathbf{0}\mathbf{.}\mathbf{0636}\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{SrH}}_{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{0636}\mathbf{\times }\mathbf{2}\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{1272}\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \end{gathered}$$

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

Step 5: Calculate the number of moles of H 2 from SrH 2

In the given reaction, 1mol of ${\mathbf{SrH}}_{\mathbf{2}}$ forms 2 mol of ${\mathbf{H}}_{\mathbf{2}}$ 

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

$$\begin{gathered} 1 \mathrm{mol} \mathrm{of} {\mathrm{SrH}}_2 = 2\mathrm{mol} \mathrm{of} {\mathrm{H}}_2 \\ 0.0636 \mathrm{mol} \mathrm{of} {\mathrm{SrH}}_2 = 0.0636 \times 2\mathrm{mol} \mathrm{of} {\mathrm{H}}_2 \\ = 0.1272\mathrm{mol} \mathrm{of} {\mathrm{H}}_2 \end{gathered}$$  

Step 6: Calculate the number of moles of from

In the given reaction, 2mol of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ forms 2 mol of ${\mathbf{H}}_{\mathbf{2}}$

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

$$\begin{gathered} 2\mathrm{mol} \mathrm{of} {\mathrm{H}}_2\mathrm{O} = 2\mathrm{mol} \mathrm{of} {\mathrm{H}}_2 \\ 0.0636 \mathrm{mol} \mathrm{of} {\mathrm{H}}_2\mathrm{O} = 0.0636 \times \frac{2}{2}\mathrm{mol} \mathrm{of} {\mathrm{H}}_2 \\ = 0.0636\mathrm{mol} \mathrm{of} {\mathrm{H}}_2 \end{gathered}$$

Step 7: Calculate the mass of H 2

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

Since the number of moles of ${\mathbf{H}}_{\mathbf{2}}$ = 0.1272mol

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

$$\begin{gathered} \mathrm{mass} \mathrm{of} {\mathrm{H}}_2 = \mathrm{moles} \mathrm{of} {\mathrm{H}}_2\times \mathrm{molar} \mathrm{mass} \mathrm{of} {\mathrm{H}}_2 \\ = 0.1272\mathrm{mol}\times 1.00784\mathrm{g}/\mathrm{mol} \\ = 0.1282 \mathrm{g} \end{gathered}$$