A heterogeneous reaction below took place in a 1000 cm³ sealed flask. C (s) + O₂ (g) ⇌ CO₂ (g). The reaction mixture initially contained 3.8 g of Carbon (C) and unknown mass of oxygen (O₂). When the equilibrium was reached, there were 0,292 mole of carbon.

Molar mass of C $= 12.01 \text{ g/mol}$.

$$\text{Initial moles of C} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.8 \text{ g}}{12.01 \text{ g/mol}} \approx 0.3164 \text{ mol}$$

At equilibrium, $0.292 \text{ mol}$ of carbon remained.

$$\text{Moles of C reacted} = \text{Initial moles of C} - \text{Equilibrium moles of C}$$ $$\text{Moles of C reacted} = 0.3164 \text{ mol} - 0.292 \text{ mol} = 0.0244 \text{ mol}$$

The reaction is: $\text{C(s)} + \text{O}_2\text{(g)} \rightleftharpoons \text{CO}_2\text{(g)}$
Let the initial moles of $\text{O}_2$ be $X$. The volume of the flask is $1000 \text{ cm}^3 = 1 \text{ dm}^3$, so moles are equal to concentrations.

Species	Initial (mol)	Change (mol)	Equilibrium (mol)
$\text{C(s)}$	0.3164	$-0.0244$	0.292
$\text{O}_2\text{(g)}$	$X$	$-0.0244$	$X - 0.0244$
$\text{CO}_2\text{(g)}$	0	$+0.0244$	0.0244

For heterogeneous reactions, solids are excluded from the $K_c$ expression.

$$K_c = \frac{[\text{CO}_2]}{[\text{O}_2]}$$

Given $K_c = 0.145$.

$$0.145 = \frac{0.0244}{X - 0.0244}$$

$$0.145(X - 0.0244) = 0.0244$$ $$0.145X - (0.145 \times 0.0244) = 0.0244$$ $$0.145X - 0.003538 = 0.0244$$ $$0.145X = 0.0244 + 0.003538$$ $$0.145X = 0.027938$$ $$X = \frac{0.027938}{0.145} \approx 0.192676 \text{ mol}$$

Molar mass of $\text{O}_2 = 2 \times 16.00 = 32.00 \text{ g/mol}$.

$$\text{Mass of O}_2 = \text{moles} \times \text{molar mass}$$ $$\text{Mass of O}_2 = 0.192676 \text{ mol} \times 32.00 \text{ g/mol} \approx 6.1656 \text{ g}$$

Rounded to two decimal places:
The initial mass of oxygen ($\text{O}_2$) placed in the flask is $\boxed{6.17 \text{ g}}$.

3.3.1 Le Chatelier's principle states that if a change of condition (such as temperature, pressure, or concentration) is applied to a system in equilibrium, the system will shift in a direction that relieves the stress and re-establishes a new equilibrium.

3.3.2 The equilibrium constant ($K_c$) only changes with temperature. Since the $K_c$ value increased, the condition that was changed is temperature.

3.3.3
The forward reaction is ENDOTHERMIC.
Explanation: An increase in the $K_c$ value indicates that the equilibrium shifted to the right, favoring the formation of products. According to Le Chatelier's principle, if the temperature of a system at equilibrium is increased, the system will shift in the direction that absorbs heat (the endothermic direction) to counteract the increase in temperature. Since the equilibrium shifted to the right (product side), the forward reaction must be endothermic.