How many grams of Al2O3 are required to react completely with 1500.0 kJ of heat? 2Al2O3(s) + 3352 kJ → 4Al(s) + 3O2(g)
Chemistry
Zae15
Question
How many grams of Al2O3 are required to react completely with 1500.0 kJ of heat? 2Al2O3(s) + 3352 kJ → 4Al(s) + 3O2(g)
2 Answer

1. User Answers Edufirst
1) Chemical reaction
2 Al2O3 (s) + 3352 KJ > 4 Al (s) + 3 O2(g)
2) State the ratio between Al2O3 and heat
2 moles of Al2O3 / 3352 kJ
3) State the proportion with the above ratio and the unknown
2 moles Al2O3 / 3352 KJ = x / 1500.0 kJ
4) Solve for x
x = 1500.0 kJ * 2 moles Al2O3 / 3352 kJ
x = 0.895 moles Al2O3.
5) Tansform moles into grams, using the molar weight
molar weight:
Al2O3: 2 * 27 g/mol + 3 * 16 g/mol = 102 g / mol
mass = molar weight * number of moles = 102 g/mol * 0.895 mol = 91.3 g.
Answer: 91.3 g 
2. User Answers castroisabel
Answer:
91.29 grams
Explanation: