To answer questions (d) and (e) regarding calcium carbonate, I need the balanced chemical equation for the reaction with hydrochloric acid and the number of moles of hydrochloric acid that reacted. Th

$$T_1 = 27^\circ C + 273 = 300 \, \text{K}$$ $$T_2 = 177^\circ C + 273 = 450 \, \text{K}$$

Charles's Law states that $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ when pressure is constant.

$$\frac{450 \, \text{cm}^3}{300 \, \text{K}} = \frac{V_2}{450 \, \text{K}}$$

$$V_2 = \frac{450 \, \text{cm}^3 \times 450 \, \text{K}}{300 \, \text{K}}$$ $$V_2 = \frac{202500}{300} \, \text{cm}^3$$ $$V_2 = 675 \, \text{cm}^3$$

The volume the gas would occupy is $\boxed{\text{675 cm}^3}$.

3. A gas occupies $4 \, \text{dm}^3$ at $-23^\circ C$ and $152 \, \text{mmHg}$. At what pressure will its volume be halved, if the temperature then is $227^\circ C$?

Initial conditions:
$V_1 = 4 \, \text{dm}^3$
$T_1 = -23^\circ C + 273 = 250 \, \text{K}$
$P_1 = 152 \, \text{mmHg}$

Final conditions:
$V_2 = \frac{V_1}{2} = \frac{4 \, \text{dm}^3}{2} = 2 \, \text{dm}^3$
$T_2 = 227^\circ C + 273 = 500 \, \text{K}$
$P_2 = ?$

The Combined Gas Law states that $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.

$$\frac{152 \, \text{mmHg} \times 4 \, \text{dm}^3}{250 \, \text{K}} = \frac{P_2 \times 2 \, \text{dm}^3}{500 \, \text{K}}$$

$$P_2 = \frac{152 \, \text{mmHg} \times 4 \, \text{dm}^3 \times 500 \, \text{K}}{250 \, \text{K} \times 2 \, \text{dm}^3}$$ $$P_2 = \frac{152 \times 4 \times 500}{250 \times 2} \, \text{mmHg}$$ $$P_2 = \frac{152 \times 4 \times 2}{2} \, \text{mmHg}$$ $$P_2 = 152 \times 4 \, \text{mmHg}$$ $$P_2 = 608 \, \text{mmHg}$$

The pressure will be $\boxed{\text{608 mmHg}}$.