0
What is the mass of Al formed when 0.500 moles Al2S3 is reduced completely with excess H2?
Wiki User
∙ 10y agoWant this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
Sodium phosphate in excess is poured into 1.00 L of a solution of aluminium sulphide at 0.25 M Calculate the mass of the precipitate formed?
Al2S3 + 2 Na3PO4 = 2 AlPO4 + 3 Na2S moles Al2S3 = 0.25 M x 1.00 L = 0.25 moles AlPO4 = 0.25 x 2 = 0.50 mass AlPO4 = 0.50 mol x 121.95 g/mol=61.0 g
How many moles of Al2S3 are in 100 g of Al2S3?
100/150.158 is 0.666 moles
What is the number of moles in 510 g of Al2S3?
510 g Al2S3 is equal to 3,396 moles.
Predict the following chemical formula for a compound between Al and S?
Al2S3
Can solid sodium chloride react with aqueous aluminium sulfide in a reaction that requires heat?
Aluminium sulfide is not soluble in water; Al2S3 is easily hydrolyzed.
What will the formula for the ionic compound formed from al3 and s2?
Al2S3====
What is the formula for a compound formed between aluminum and sulfur atoms?
Al2S3
Is sulfur reactive to aluminum?
Yes, and the compound Al2S3 (aluminium sulfide) is formed.
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.
Sodium phosphate in excess is poured into 1.00 L of a solution of aluminium sulphide at 0.25 M Calculate the mass of the precipitate formed?
Al2S3 + 2 Na3PO4 = 2 AlPO4 + 3 Na2S moles Al2S3 = 0.25 M x 1.00 L = 0.25 moles AlPO4 = 0.25 x 2 = 0.50 mass AlPO4 = 0.50 mol x 121.95 g/mol=61.0 g
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 )
How do you balance Al plus S equals Al2S3?
2Al+3S--->Al2S3
How many moles of Al2S3 are in 100 g of Al2S3?
100/150.158 is 0.666 moles
What is the number of moles in 510 g of Al2S3?
510 g Al2S3 is equal to 3,396 moles.
Chemical formula for aluminum sulfide?
S2Ur5 NO this is wrong! There isn't even an element that is Ur wow! The answer is Al2S3
Is Al2S3 soluble?
Yes
What is the formula unit of aluminum sulfide?
The chemical formula for aluminium sulfide is Al2S3.
