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${}^{\mathbf{C}\mathbf{a}}$ = 4.20 g.

The molar mass of${}^{\mathbf{Ca}}$=40.078 g/mol

Thus, thenumber of moles${}^{\mathbf{Ca}}$is:

role="math" localid="1657095204157" $$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{Ca}=\frac{\mathrm{mass}\mathrm{of}\mathrm{Ca}}{\mathrm{molar}\mathrm{mass}\mathrm{of}\mathrm{Ca}} \\ =\frac{4.20\mathrm{g}}{40.078\mathrm{g}/\mathrm{mol}} \\ =0.1048\mathrm{mol} \end{gathered}$$

The mass of${}^{{\mathbf{O}}_{\mathbf{2}}}$= 2.80 g.

The molar mass of${}^{{\mathbf{O}}_{\mathbf{2}}}$=31.998g/mol

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

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{O}}_2=\frac{\mathrm{mass}\mathrm{of}{\mathrm{O}}_2}{\mathrm{molar}\mathrm{mass}\mathrm{of}{\mathrm{O}}_2} \\ =\frac{2.80\mathrm{g}}{31.998\mathrm{g}/\mathrm{mol}} \\ =0.875\mathrm{mol} \end{gathered}$$

In the reaction, 2 mol o ${}^{\mathbf{C}\mathbf{a}}$reacts with 1mol of${}^{{\mathbf{O}}_{\mathbf{2}}}$Thus,

role="math" localid="1657095926696" $$\begin{gathered} 2\mathrm{mol}\mathrm{of}\mathrm{Ca}=1\mathrm{mol}\mathrm{of}{\mathrm{O}}_2 \\ 0.1048\mathrm{mol}\mathrm{of}\mathrm{Ca}=0.1048×\frac{1}{2}{\mathrm{molofO}}_2 \\ =0.0524\mathrm{molof}{\mathrm{O}}_2 \end{gathered}$$

Therefore,role="math" localid="1657095760078" ${}^{{\mathrm{O}}_2}$present in access in the reaction, hence${}^{\mathbf{C}\mathbf{a}}$is a limiting reagent.

In the given reaction,2mol of${}^{\mathbf{C}\mathbf{a}}$forms 2mol of${}^{\mathbf{C}\mathbf{a}\mathbf{O}}$

Thus, the number of moles of is

role="math" localid="1657096177880" $$\begin{gathered} 2\mathrm{mol}\mathrm{of}\mathrm{Ca}=2\mathrm{mol}\mathrm{of}Ca\mathrm{O} \\ 0.1048\mathrm{mol}\mathrm{of}\mathrm{Ca}=0.1048×\frac{2}{2}\mathrm{molof}CaO \\ =0.1048\mathrm{molof}Ca\mathrm{O} \end{gathered}$$

In the given reaction,1 mol of ${}^{{\mathbf{O}}_{\mathbf{2}}}$forms 2 mol of role="math" localid="1657096151341" ${}^{\mathbf{C}\mathbf{a}\mathbf{O}}$

Thus, the number of moles of${}^{\mathbf{CaO}}$is

$$\begin{gathered} 1\mathrm{mol}\mathrm{of}{O}_2=2\mathrm{mol}\mathrm{of}Ca\mathrm{O} \\ 0.0875\mathrm{mol}\mathrm{of}{O}_2=0.1048×2\mathrm{mol}\mathrm{of}CaO \\ =0.175\mathrm{mol}\mathrm{of}Ca\mathrm{O} \end{gathered}$$

The molar mass of${}^{\mathbf{CaO}}$=56.08g/mol

Since the number of moles of${}^{\mathbf{CaO}}$= 0.1048 mol

Thus, themassof${}^{\mathbf{CaO}}$is:

$$\begin{gathered} \mathrm{mass}\mathrm{of}\mathrm{Ca}=\mathrm{moles}\mathrm{of}\mathrm{CaO}×\mathrm{molar}\mathrm{mass}\mathrm{of}\mathrm{CaO} \\ =0.1048\mathrm{mol}×56.08\mathrm{g}/\mathrm{mol} \\ =5.88\mathrm{g} \end{gathered}$$