Q73P - Chemistry: Molecular Nature Of Matter And Change

Question

Many metals react with oxygen gas to form the metal oxide. For example, calcium reacts as follows: 

$\mathbf{2}\mathbf{Ca}\left(s\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}\mathbf{CaO}\left(s\right)$

You wish to calculate the mass of calcium oxide that can be prepared from 4.20 g of  and 2.80 g of ${\mathbf{O}}_{\mathbf{2}}$

(a) How many moles of CaO can be produced from the given mass of Ca ?

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

(c) Which is the limiting reactant? 

(d) How many grams of CaO can be produced?

Step-by-Step Solution

Answer

1. The number of moles of CaO  produced from Ca is 0.1048 mol.
2. The number of moles of CaO  produced from ${\mathbf{O}}_{\mathbf{2}}$ is 0.175 mol.
3. CaO is the limiting reactant.
4. The mass of CaO produced is 5.88 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 C a

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

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

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

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

Step 3: Calculate the number of moles of O 2

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

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

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

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

Step 4: Determine the Limiting reagent of the reaction

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

$$\begin{gathered} 2 \mathrm{mol} \mathrm{of} \mathrm{Ca} = 1\mathrm{mol} \mathrm{of} {\mathrm{O}}_2 \\ 0.1048\mathrm{mol} \mathrm{of} \mathrm{Ca} =0.1048\times \frac{1}{2}{\mathrm{molofO}}_2 \\ =0.0524\mathrm{molof} {\mathrm{O}}_2 \end{gathered}$$

Therefore, ${}^{{\mathrm{O}}_2}$ present in access in the reaction, hence${}^{\mathbf{C}\mathbf{a}}$ is a limiting reagent.

Step 5: Calculate the number of moles of C a O from C a

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

$$\begin{gathered} 2 \mathrm{mol} \mathrm{of} \mathrm{Ca} = 2\mathrm{mol} \mathrm{of}Ca \mathrm{O} \\ 0.1048\mathrm{mol} \mathrm{of} \mathrm{Ca} =0.1048\times \frac{2}{2}\mathrm{molof}CaO \\ =0.1048\mathrm{molof} Ca\mathrm{O} \end{gathered}$$

Step 6: Calculate the number of moles of C a O from O 2

In the given reaction, 1 mol of ${}^{{\mathbf{O}}_{\mathbf{2}}}$ forms 2 mol of ${}^{\mathbf{C}\mathbf{a}\mathbf{O}}$

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

$$\begin{gathered} 1 \mathrm{mol} \mathrm{of} O_2 = 2\mathrm{mol} \mathrm{of}Ca \mathrm{O} \\ 0.0875\mathrm{mol} \mathrm{of} O_2 =0.1048\times 2\mathrm{mol} \mathrm{of}CaO \\ =0.175\mathrm{mol} \mathrm{of} Ca\mathrm{O} \end{gathered}$$

Step 7: Calculate the mass of CaO

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

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

Thus, the mass of${}^{\mathbf{CaO}}$ is:

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