Q. 7.93 - An Introduction to General, Organic, and Biological Chemistry

Question

How many grams are in $0.150$ mole of each of the following? ($7.3$)

a. $\mathrm{K}$

b. ${\mathrm{Cl}}_2$

c. $N{a}_{2}C{O}_3$

Step-by-Step Solution

Answer

(a) The mass of $K$ in $0.150$ mole is $5.87$ g

(b) The mass of $C{l}_2$ in style="max-width: none; vertical-align: -4px;" $0.150$ mole is $10.6$ g

(c) The mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ in $0.150$ mole is $15.9$ g

Part (a) Step 1: Given information

We need to find the mass of given element i.e. $K$

Part (a) Step 2: Explanation

We know that

Molar mass of $\mathrm{K}$ is $39 \mathrm{g}/\mathrm{mole}$

And a mole of the element is equal to molar mass

Therefore,

$$\begin{gathered} 1 \mathrm{mole}=39 \mathrm{g} \\ 0.15 \mathrm{mole}=\frac{39 \mathrm{g}\times 0.150 \cancel{\mathrm{mole}}}{\cancel{\mathrm{mole}}} =5.87 \mathrm{g} \end{gathered}$$

Part (b) Step 1: Given information

We need to find the mass of given compound i.e. ${\mathrm{Cl}}_2$

Part (b) Step 2: Explanation

We know that

Molar mass of ${\mathrm{Cl}}_2$ is $71 \mathrm{g}/\mathrm{mole}$

And a mole of the element is equal to molar mass

Therefore,

$$\begin{gathered} 1 \mathrm{mole}=71\mathrm{g} \\ 0.150 \mathrm{mole}=\frac{71 \mathrm{g} \times 0.150 \cancel{\mathrm{mol}}}{\cancel{\mathrm{mol}}}=10.6 \mathrm{g} \end{gathered}$$

Part (c) Step 1: Given information

We need to find the mass of given compound i.e. ${\mathrm{Na}}_2{\mathrm{CO}}_3$

Part (c) Step 2: Explanation

We know that

Molar mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ is $106 \mathrm{g}/\mathrm{mole}$

And a mole of the element is equal to molar mass

Therefore,

$$\begin{gathered} 1 \mathrm{mole}=106 \mathrm{g} \\ 0.150 \mathrm{mole}=\frac{106 \mathrm{g} \times 0.150 \cancel{\mathrm{mol}}}{\cancel{\mathrm{mol}}}=15.9 \mathrm{g} \end{gathered}$$