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Q: How do you balance Al plus S equals Al2S3?
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What is the chemical formula for the ionic compound formed from aluminum and sulfur?

Al2S3. Obviously, the 2 and 3 are subscript. The reason this happens is because of the bonding ratio: there are 3 S ions required to balance 2 Al ions.

Chemical formula for aluminum sulfide?

S2Ur5 NO this is wrong! There isn't even an element that is Ur wow! The answer is Al2S3

How many metal atoms are in Al2S3?

Each molecule of Al2S3 is composed of one Al3+ ion. Since a mole is equal to 6.022 x 1023 molecules, there would be 6.022 x 1023 Al3+ ions in one mole of Al2S3.

What is the oxidation number of Al2S3?

The oxidation number of Al is +III and S is -II.

What is the oxidation number of sulfur in Al2S3?

S has an oxidation number of -2. Al is +3.

Predict the following chemical formula for a compound between Al and S?


How do you balance the equation Al plus O2 yields Fe2O3?

Al + O2 will never yield Fe2O3.

What is the oxidation half-reaction of 3Ag Al 3Ag Al3?

Al and AgNO3. Al has a 3plus charge plus 3e negative. 3(Ag positive plus le negative equals Ag). Al plus 3Ag plus is equal to Al with a 3 plus charge plus 3Ag.

How do you balance out Al plus Cl Cu?

The chemical equation is:2 Al +3 CuCl2 = 3 Cu + 2 AlCl3

How many grams of Al2S3 can be formed from the reaction of 108.00 grams of Al with 5.00 grams of S?

The answer i got is 12.33 grams of Al2S3. Below i will try to show the steps i used: n=moles m=mass (grams) M= molecular weight (from periodic table) 2Al + 3S --> Al2S3 nAl= m/M = 9/13 = 0.692307 moles of Al nS= m/M = 8/16 = 0.5 moles of S Limiting Reagent: 1.5 moles of S required for every mole of Al (3:2 ratio) This means that there should be about 1.04 moles of S to completely use up all the Al. Since there is less than this amount of S present (only 0.5 moles), S is the limiting reagent and should be used in the mole calculation of the product Amount of product: 0.5 moles S x (1 mol Al2S3/ 3 mol S) = 0.16666 moles of Al2S3 nAl2S3 = m/M Molecular weight of Al2S3 = (2 x 13) + (3 x 16) = 74 0.16666 moles Al2S3 = m/74 m = 74 x 0.16666 = 12.33 grams of Al2S3 Therefore, 12.33 grams of Al2S3 is formed in this reaction ( hopefully this is right :P )

What is the coeficient needed for Al to balance Al plus O2 - Al2O3?

The balanced chemical equation for the reaction between Al and O2 is: 4Al + 3O2 -> 2Al2O3 Therefore, the coefficient needed for Al to balance the equation is 4.

How many grams of aluminium hydroxide are obtained from 14.2g of aluminium sulfide?

Moles Al2S3 = 14.2 g / 150.16 g/mol =0.9456 the rario between Al2S3 and Al(OH)3 is 1 : 2 moles Al(OH)3 = 2 x 0.9456 =0.1891 Mass Al(OH)3 = 0.1891 mol x 78 g/mol =14.75 g