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Determine the Theoritical yield of H2S in moles if 4.0 mol Al2S3 and 4.0 mol H2O are reacted?
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∙ 11y ago
6moles
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∙ 11y ago
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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 )
If 10 moles of A3 are reacted with excess B2 how many moles of AB will be produced 2A3 plus 3B2 6AB?
30 moles
2A3 plus 3B2 6AB If 10 moles of A3 are reacted with excess B2 how many moles of AB will be produced?
The answer is 30 moles.
An excess of Mg is reacted with 1 mole of HCl dissolved in water- determine the mass of hydrogen produced?
Mg + 2HCl => MgCl2 + H2 2 moles HCl : 1 mole H2. 2.02 x .5 moles = 1.01 1.01g H2
Identify the number of moles of excess reagent left over when 4.821 moles of ironIII oxide are reacted with 5.591 moles of aluminum in the thermite reaction?
2.026 mole ironIII oxide
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.
How many moles of N2 reacted if 0.60 mol of NH3 is produced?
0,3 moles of nitrogen reacted.
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 )
Which is the limiting reactant when 3.00 moles of calcium are reacted with 8.00 moles of water in the following equation?
H2o
If 10 moles of A3 are reacted with excess B2 how many moles of AB will be produced 2A3 plus 3B2 6AB?
30 moles
2A3 plus 3B2 6AB If 10 moles of A3 are reacted with excess B2 how many moles of AB will be produced?
The answer is 30 moles.
An excess of Mg is reacted with 1 mole of HCl dissolved in water- determine the mass of hydrogen produced?
Mg + 2HCl => MgCl2 + H2 2 moles HCl : 1 mole H2. 2.02 x .5 moles = 1.01 1.01g H2
Identify the number of moles of excess reagent left over when 4.821 moles of ironIII oxide are reacted with 5.591 moles of aluminum in the thermite reaction?
2.026 mole ironIII oxide
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
For the reaction 2H2 plus O2 equals 2H2O how much water is produced when 2.5 moles of hydrogen react completely How do you do the problem?
2H2 + O2 ---------------> 2H2O for every 2 moles of hydrogen that reacts, 2 moles of water are produced, thus a 1:1 ratio of water produced to hydrogen reacted. So:- 2.5 moles of hydrogen reacted will produce 2.5 moles of water
How many moles of O2 must react with elemental chlorine to produce 69.1 moles of dichlorine heptoxide?
2Cl2 + 7O2 --> 2Cl2O7So for every 2 moles of dichlorine heptoxide formed, 7 moles of oxygen are reacted.69.1 moles Cl2O7 * (7 moles O2 / 2 moles Cl2O7) = 241.85 moles O2
