∆H for given reaction -847kJ
- As it's negative reaction is exothermic
So
2 mol of Al releases 847KJ heat
4 mol Al releases
∆H=-1694KJ
If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.
In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.
So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC