Question

The standard heat of combustion of $Al$ is $-837.8kJmo{l}^{-1}$ ${25}^{\circ }C$ which of the following releases $250kcal$ of heat?

• A. The reaction of 0.624 mol of Al
• B. The formation of 0.624 mol of Al₂O₃
• C. The reaction of 0.312 mol of Al
• D. The formation of 0.150 mol of Al₂O₃

Answer 1

The correct option is B

The formation of 0.624 mol of Al₂O₃

$Al+\frac{3}{4}{O}_2\to \frac{1}{2}A{l}_2{O}_3$ $\Delta H=-837.8kJ$

released energy = $250kcal$ = $250÷4.2=1050kJ$

$\because 837.8kJ$ energy is released in formation of 0.5 mole $A{l}_2{O}_3$

$\therefore$ $1050kJ$ energy released = $\frac{0.5}{837.8}\times 1050$

= $0.624mole$