The alumina refinery processes 3000 tons of ore containing 54% Al2O3 daily. Al2O3 is in the ore in the form

| August 17, 2021 | education

The alumina refinery processes 3000 tons of ore containing 54% Al2O3 daily. Al2O3 is in the ore in the form of the gibbsite mineral Al2O3 * 3H2O. The ore processing technology provides for the transfer of Al2O3 into solution by the reaction Al2O3 * 3H2O + 2NaOH – 2NaAl (OH) 4. Determine the daily requirement of NaOH alkali, subject to 88% extraction of Al2O3 in solution.

1. Let’s determine the Al2O3 content in 3000 tons of ore:

3000 * 54/100 = 1620 t.

2. Determine the mass of Al2O3 that will react:

1620 * 88/100 = 1425.6 t.

3. The reaction equation has the form:

Al2O3 * 3H2O + 2NaOH = 2NaAl (OH) 4.

Molar mass of Al2O3 = 102 g (27 * 2 + 16 * 3 = 102).

Molar mass of 2NaOH = 80 g (2 * (23 + 16 + 1) = 80).

4. We make up the proportion:

102 – 80

1425.6 – x.

x = 1425.6 * 80/102 = 1118.12 t.

Answer: the daily requirement for alkali is 1118, 12 tons.



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