A) The reaction of 0.624 mol of Al
B) The formation of 0.624 mol of \[A{{l}_{2}}{{O}_{3}}\]
C) The reaction of 0.312 mol of \[Al\]
D) The formation of 0.150 mol of \[A{{l}_{2}}{{O}_{3}}\]
Correct Answer: B
Solution :
| [b] \[Al+\frac{3}{4}{{O}_{2}}\xrightarrow{{}}\frac{1}{2}A{{l}_{2}}{{O}_{3}};\,\,\Delta H=-\text{ }837.8\text{ }kJ\] |
| Realize energy \[=250\text{ }kcal=250\times 4.2=1050\text{ }kJ\] |
| Since 837.8 kJ energy is realized in formation of 0.5 mol \[A{{l}_{2}}{{O}_{3}},\] therefore 1050 kJ energy realized \[\frac{0.5}{837.8}\times 1050=0.624\,mol\] |
You need to login to perform this action.
You will be redirected in
3 sec
