How many grams of al2o3 are required to react completely with 1500.0 kj of heat? 2al2o3(s) + 3352 kj → 4al(s) + 3o2(g)

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.

91.3 g

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