Final answer:
To calculate the milliequivalents of aluminum and carbonate ions in 100 mg of aluminum carbonate, first convert the mass of aluminum carbonate to moles and then use the valence of the ions to find the milliequivalents. There are approximately 2.56 milliequivalents of each aluminum and carbonate ions present in 100 mg of aluminum carbonate.
Step-by-step explanation:
To calculate the number of milliequivalents of aluminum and carbonate ions in 100 mg of aluminum carbonate, we need to know the molar mass of aluminum carbonate and understand that a milliequivalent is equal to the molarity of the ion times its valence divided by 1000. Aluminum carbonate has the chemical formula Al2(CO3)3, and from periodic table data, we can determine its molar mass is approximately 233.99 g/mol (summing the molar masses of 2 aluminum atoms, 3 carbon atoms, and 9 oxygen atoms).
Firstly, we will convert the mass of aluminum carbonate to moles:
- 100 mg of aluminum carbonate is equal to 0.1 g.
- The number of moles of aluminum carbonate is 0.1 g / 233.99 g/mol = 4.27 x 10-4 moles.
Since aluminum has a valence of +3 and there are 2 aluminum ions per formula unit of aluminum carbonate, the total milliequivalents of aluminum would be:
- (4.27 x 10-4 moles) x 3 (valence of Al3+) x 2 (Al ions per formula unit) x 1000 = 2.56 mEq.
Similarly, the carbonate ion has a valence of -2 and there are 3 carbonate ions per formula unit of aluminum carbonate. The milliequivalents of carbonate ions would be:
- (4.27 x 10-4 moles) x 2 (valence of CO32-) x 3 (CO3 ions per formula unit) x 1000 = 2.56 mEq.
In conclusion, 100 mg of aluminum carbonate consists of 2.56 milliequivalents of aluminum ions and 2.56 milliequivalents of carbonate ions.