How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
Al + O2 = Al2O3
Using PV=nRT
n=PV/RT
n=(97.3×15)/8.31×294
n=0.6
Mass =mole×molar mass
Mass =0.6×32
Mass=19.2g
If 32g of O2 produces 102g of Al2O3
Then 19.2g will produce Xg of Al
X=61.2g of Al2O3.
Comments
Leave a comment