91Ó°ÊÓ

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.

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

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

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

Thus, thenumber of molesis ${\mathbf{SrH}}_{\mathbf{2}}$:

role="math" localid="1657097769684" $$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{mass}{\mathrm{SrH}}_2}{\mathrm{Molar}\mathrm{mass}{\mathrm{SrH}}_2} \\ =\frac{5.70\mathrm{g}}{89.64\mathrm{g}/\mathrm{mol}} \\ =0.0636\mathrm{mol} \end{gathered}$$

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

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

Thus, thenumber of moles${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$is:

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

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

role="math" localid="1657098084066" $$\begin{gathered} \mathbf{1}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{SrH}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ \mathbf{0}\mathbf{.}\mathbf{0636}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{SrH}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{0636}\mathbf{×}\mathbf{2}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{1272}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \end{gathered}$$

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

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

role="math" localid="1657098559216" $$\begin{gathered} 1\mathrm{mol}\mathrm{of}{\mathrm{SrH}}_2=2\mathrm{mol}\mathrm{of}{\mathrm{H}}_2 \\ 0.0636\mathrm{mol}\mathrm{of}{\mathrm{SrH}}_2=0.0636×2\mathrm{mol}\mathrm{of}{\mathrm{H}}_2 \\ =0.1272\mathrm{mol}\mathrm{of}{\mathrm{H}}_2 \end{gathered}$$

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

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

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

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

Thus, themassof${\mathbf{H}}_{\mathbf{2}}$is:

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