A method for recycling scrap aluminum converts it to alum [potassium aluminum sulfate dodecahydrate, KAl(SO4)2·12H2O].
What is the maximum amount of alum (in metric tons) that can be prepared from 3.12 metric tons of scrap aluminum (1 metric ton = 1000 kg)?
This is a simple stoichiometry problem. Here is a worked example of one. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the maximum amount of alum that can be prepared from 3.12 metric tons of scrap aluminum, we need to consider the stoichiometry of the reaction between scrap aluminum and alum.
The balanced chemical equation for the reaction is:
2Al + 2KOH + 4H2SO4 + 22H2O -> 2KAl(SO4)2·12H2O + 3H2
From the equation, we can see that two moles of aluminum (Al) react to produce one mole of alum (KAl(SO4)2·12H2O).
1 mole of aluminum weighs 26.98 grams. Therefore, we can calculate the moles of aluminum present in 3.12 metric tons of scrap aluminum as follows:
3.12 metric tons * 1000 kg/ton * 1000 g/kg / 26.98 g/mole = 115,897.28 moles of aluminum
Since the stoichiometry of the reaction is 2 moles of aluminum to 1 mole of alum, we can divide the moles of aluminum by 2 to get the moles of alum:
115,897.28 moles of aluminum / 2 = 57,948.64 moles of alum
Now, to convert the moles of alum to metric tons, we calculate the molar mass of alum:
KAl(SO4)2·12H2O:
- Potassium (K) atomic mass = 39.10 g/mol
- Aluminum (Al) atomic mass = 26.98 g/mol
- Sulfur (S) atomic mass = 32.07 g/mol
- Oxygen (O) atomic mass = 16.00 g/mol (12 O atoms in the formula)
- Hydrogen (H) atomic mass = 1.008 g/mol (24 H atoms in the formula)
Total molar mass = (39.10 g/mol) + (26.98 g/mol) + 2(32.07 g/mol) + 12(16.00 g/mol) + 12(1.008 g/mol) = 474.4 g/mol
Now, we convert moles to metric tons:
57,948.64 moles of alum * (474.4 g/mol) / (1000 kg/ton) / 1000 = 27.50 metric tons of alum
Therefore, the maximum amount of alum that can be prepared from 3.12 metric tons of scrap aluminum is 27.50 metric tons.