21 Ҳангоми сӯхтани 216 г алюминий мувофиқи реаксияи термохимиявии 2Al + 1,5O₂ = Al₂O₃ + 1640 кҶ чанд кҶ гармӣ хорич мешавад?

Solution:

The reaction is given as: \( 2\text{Al} + 1.5\text{O}_2 = \text{Al}_2\text{O}_3 + 1640 \text{ kJ} \). This means that when 2 moles of aluminum react, 1640 kJ of heat are released.

First, calculate the molar mass of aluminum (Al).

Molar mass of Al = 26.98 g/mol.

Now, calculate the number of moles of aluminum in 216 g:

\[ n(\text{Al}) = \frac{\text{mass}}{\text{molar mass}} = \frac{216 \text{ g}}{26.98 \text{ g/mol}} \approx 8.01 \text{ mol} \]

From the stoichiometry of the reaction, 2 moles of Al release 1640 kJ of heat.

We can set up a proportion to find the heat released for 8.01 moles of Al:

\[ \frac{1640 \text{ kJ}}{2 \text{ mol Al}} = \frac{x \text{ kJ}}{8.01 \text{ mol Al}} \]

Solving for x:

\[ x = \frac{1640 \text{ kJ} \times 8.01 \text{ mol}}{2 \text{ mol}} \approx \frac{13136.4}{2} \approx 6568.2 \text{ kJ} \]

Rounding to a more appropriate number of significant figures based on the input (216 g has 3 significant figures, 1640 kJ has 4):

\[ x \approx 6570 \text{ kJ} \]

Answer: 6570 кҶ