The Atomic Mass of Al is 27.0
Amount of Al in 5.39g pure sample = 5.39/27.0 = 0.200mol
1,46 moles of aluminum fluoride contain 35,16848.10e23 atoms.
4,12 grams aluminum sulfate is equivalent to 0,012 moles (for the anhydrous salt).
27g
I think you mean the potassium aluminum silicate form of mica. KAlSi3O8 2.80 KAlSi3O8 (1 mole KAlSi3O8/278.35 grams)(1 mole Al/1 mole KAlSi3O8) = 0.010059 moles aluminum ------------------------------------( 1.01 X 10^-3 moles )
9 grams aluminum (1 mole Al/26.98 grams) = 0.3 moles aluminum ==============
1,99 grams of aluminum is equal to 0,0737 moles.
10 grams aluminum (1 mole Al/26.98 grams) = 0.37 moles of aluminum ---------------------------------
There are 6 moles of sulfur present in 3 moles of aluminum sulfate, because aluminum sulfate has a 2:3 ratio of aluminum to sulfur.
When 4 moles of aluminum react with an excess of chlorine gas, 4 moles of aluminum chloride are produced. This is because the balanced chemical equation for the reaction is: 2Al + 3Cl2 -> 2AlCl3 This means that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride, so 4 moles of aluminum will produce 4 moles of aluminum chloride.
1,46 moles of aluminum fluoride contain 35,16848.10e23 atoms.
To determine how many moles of aluminum are produced from 33 grams, divide the given mass by the molar mass of aluminum, which is approximately 26.98 g/mol. So, 33 g / 26.98 g/mol ≈ 1.22 moles of aluminum are produced.
When aluminum oxide decomposes, it produces 2 moles of aluminum and 3 moles of oxygen for every mole of aluminum oxide. Therefore, for 26.5 moles of aluminum oxide decomposed, 3 * 26.5 = 79.5 moles of oxygen are produced.
When 4 moles of aluminum react with an excess of chlorine gas (Cl2), 4 moles of aluminum chloride are produced because the balanced chemical equation for this reaction is: 2 Al + 3 Cl2 -> 2 AlCl3 Since the mole ratio between aluminum and aluminum chloride is 2:2, it means that for every 2 moles of aluminum, 2 moles of aluminum chloride are produced.
To determine the number of moles of aluminum present, we need to first determine the molar mass of aluminum, which is approximately 26.98 g/mol. We can then use the formula: moles = mass / molar mass. Plugging in the values, we get moles = 15 g / 26.98 g/mol ≈ 0.56 moles of aluminum.
There would be 4.38 moles of fluoride ions in 1.46 moles of aluminum fluoride, as the formula for aluminum fluoride is AlF3 with three fluoride ions per molecule of aluminum fluoride.
4,12 grams aluminum sulfate is equivalent to 0,012 moles (for the anhydrous salt).
3.2 moles of water (H2O)