The Mole and Stoichiometry: Question 4

Worked solution

Part (a): Moles of cobalt(II) carbonate

First find the relative formula mass of $\text{CoCO}_3$:

$M_r(\text{CoCO}_3) = A_r(\text{Co}) + A_r(\text{C}) + 3\times A_r(\text{O}) = 59+12+(3\times16)=59+12+48=119$

Use $\text{moles} = \dfrac{\text{mass}}{\text{molar mass}}$:

$\text{moles of CoCO}_3 = \frac{23.8}{119} = 0.2\ \text{mol}$

Part (b): Moles and mass of cobalt(II) oxide

The balanced equation, $\text{CoCO}_3\text{(s)} \rightarrow \text{CoO(s)} + \text{CO}_2\text{(g)}$, has a mole ratio of $1:1:1$ between all three substances.

So $0.2\ \text{mol}$ of $\text{CoCO}_3$ produces $0.2\ \text{mol}$ of $\text{CoO}$.

Find the molar mass of $\text{CoO}$:

$M_r(\text{CoO}) = 59+16=75$

Convert moles to mass using $\text{mass} = \text{moles}\times\text{molar mass}$:

$\text{mass of CoO} = 0.2\times75=15\ \text{g}$

Part (c): Volume of carbon dioxide at r.t.p.

By the same $1:1:1$ mole ratio, $0.2\ \text{mol}$ of $\text{CO}_2$ gas is also produced.

At room temperature and pressure, r.t.p., $1\ \text{mol}$ of any gas occupies $24\ \text{dm}^3$:

$\text{volume of CO}_2 = 0.2\times24=4.8\ \text{dm}^3$

Final answers

• (a) Moles of $\text{CoCO}_3 = \boxed{0.2}\ \text{mol}$
• (b) Moles of $\text{CoO} = 0.2\ \text{mol}$; mass of $\text{CoO} = \boxed{15}\ \text{g}$
• (c) Volume of $\text{CO}_2$ at r.t.p. $= \boxed{4.8}\ \text{dm}^3$