Chemistry

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) 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
  • Answer:

    91.29 grams

    Explanation:

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