How many grams of Al¿OÀ are required to react completely with 1500.0 kJ of heat? 2Al¿OÀ(s) + 3352 kJ d 4Al(s) + 3O¿(g)

Question

abmon98 · Answer

2Al2O3(s) + 3352 kJ --> 4Al(s) + 3O2(g)
2 moles of Al2O3 react with 3352 KJ, so in order for 1 mole of Al2O3 to react we need 3352/2 of heat. Suppose that 1676 is the molar Heat.
1 mole of Al2O3 --->1676 KJ
x mole of Al2O3 --->1500KJ
1500*1/1676=0.89 moles
Number of Moles=Mass(g)/Molar Mass(g/mol)
Molar Mass Of Aluminium Oxide: Aluminium: 27grams O:16 grams 
Al2O3: (27*2)+(16*3)=102 grams 
Mass=Molar Mass(g/mol)*Number of Moles
0.89*102=90.78 grams