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510 g Al2S3 is equal to 3,396 moles.
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∙ 7y ago
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How many moles of Al2S3 are in 100 g of Al2S3?
100/150.158 is 0.666 moles
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 is a number of moles of an element determined from a known mass?
The number of moles is mass in g/molar mass in g.
What is the number of moles in 0.025 g NH42Cr2O7?
The number of moles 9,92.10e-5.
What is the number of moles in 432 g Mg (OH)2?
The number of moles in 432 g Mg (OH)2 is 7,407.
How many moles of Al2S3 are in 100 g of Al2S3?
100/150.158 is 0.666 moles
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
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 )
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 is a number of moles of an element determined from a known mass?
The number of moles is mass in g/molar mass in g.
If you have g of beryllium how many moles is this equivalent to?
The formula is: number of moles = g Be/9,012.
What is the number of moles in 0.025 g NH42Cr2O7?
The number of moles 9,92.10e-5.
What is the number of moles in 15 g of As H3?
0.0027 moles.
How many moles of SO2 are present in 250.0 g?
Number of moles is determined by dividing molar mass into the number of grams. SO2 has a molar mass of 64.066 g. To find the number of moles in 250.0 g of SO2, divide 250.0 g by 64.066 g. This gives you just over 3.9 moles.
What is the number of moles if you have 22.99 g of C28H44O?
22.99 g of C28H44O is equal to 0,058 moles.
What is the number of moles in 432 g Mg (OH)2?
The number of moles in 432 g Mg (OH)2 is 7,407.
Find the number of moles in 607 g krypton?
7.24 moles.
