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100/150.158 is 0.666 moles
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What is the number of moles in 510 g of Al2S3?
510 g Al2S3 is equal to 3,396 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 many moles are in 23.34 g of NaCl?
The answer is 0,4 moles.
How many moles are in 978 g Ca?
978 g calcium contain 24,4 moles.
How many moles are in 14.84 g Mg?
14,84 g magnesium are equivalent to 0,61 moles.
What is the number of moles in 510 g of Al2S3?
510 g Al2S3 is equal to 3,396 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 )
How many moles are in 100g of KMnO4?
1 mol = 118.94 1 mol / 118.94 = 1.70 / x G = 202.10g
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
What is the number of moles in 100 g MgCO3?
100 grams MgCO3 (1mol MgCO3/84.32g) = 1.19 moles of MgCO3
How many atoms of aluminum are there in a sample of aluminum sulfide with a mass of 100 g?
Aluminum sulfide has a molar mass of 150.16 grams per mole. This means there are 0.666 moles present, or 4.01 E23 molecules. Each molecule of Al2S3 has 2 aluminum atoms, so there are 8.02 E23 atoms of aluminum present.
How many moles does 3.80 g Zn have?
3,80 g Zn have 0,058 moles.
If you have g of beryllium how many moles is this equivalent to?
The formula is: number of moles = g Be/9,012.
How many moles of argon are in 24.7 g of argon?
Moles = weight in g / atomic weight. So moles in 24.7 g of Ar = 24.7 / 39.948 = 0.62 moles
How many moles are in 19.2 g of CS2?
The answer is 0,253 moles.
How many moles are in 23.34 g of NaCl?
The answer is 0,4 moles.
