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Ammonia-NH3
2N+3H2=2NH3
2moles of Nitrogen produced 2moles of Ammonia
(2*14)g of Nitrogen produced (2*17)g of Ammonia
28g of Nitrogen produced 34g of Ammonia
34g of Ammonia is produced by 28g of Nitrogen
0.034kg of Ammonia is produced by 0.028kg of Nitrogen
91.3kg of Ammonia will be produced by 0.028*91.3/0.034
91.3kg of Ammonia will be produced by 75.19kg of Nitrogen
FOR HYDROGEN: 3moles of H2 produces 2moles of NH3
(2*3)g H2 produces 2*17g NH3
6g hydrogen produces 34g ammonia
0.006kg hydrogen produces o.o34kg ammonia
91.3kg ammonia will be produced by 91.3*0.006/.034=16.11kg of Hydogen
Therefore, 75.19kg of Nitrogen and 16.11kg of Hydrogen will produce 91.3kg of Ammonia
What else can I help you with?
How many moles of nitrogen and hydrogen are needed to get 10 moles of ammonia?
The reaction of nitrogen with hydrogen to form ammonia is: N2 +3H2 = 2NH3 Therefore to make 10 moles of ammonia you need 5 moles N2 and 15 moles H2
How many moles of hydrogen are required to produce 18.00 moles of ammonia?
N2 + 3H2 -> 2NH3 The stoichiometric equation (or balanced equation) for the formation of ammonia from this we can read off the mole ratio between hydrogen and ammonia; 3M H2 needed to produce 2M NH3 times each by 9 (so the ratio remains the same and 18M NH3 is formed) 27M H2 needed to produce 18M NH3
How many grams of nitrogen are needed to produce 660 grams of ammonia?
The equation for the formation of ammonia from nitrogen and hydrogen gases at standard temperature and pressure is N2 + 3 H2 -> 2 NH3. The gram molecular mass of ammonia is 17.03 and the gram molecular mass of divalent nitrogen is 28.0134. Therefore, designating the unknown mass of nitrogen needed as p, the proportion 660/p equals (2 X 17.03)/28.0134 is valid. This expression is algebraically equivalent to 660 = (34.06)p/28.0134, for which p = 543 grams, rounded to the justified number of significant digits, limited by the three significant digits specified by the given number 600.
What mass of nitrogen is needed to make 34 grams of ammonia?
To find the mass of nitrogen needed to make ammonia, first determine the molar mass of ammonia (NH3) which is 17 g/mol. Since there is one nitrogen atom in ammonia, the nitrogen mass is 14 g/mol. To make 34 grams of ammonia, you would need 14 grams of nitrogen.
In order to get 1 mole of ammonia how many moles of nitrogen is needed?
Each mole of ammonia requires one mole of nitrogen atoms. However, the nitrogen in the air occurs as diatomic molecules; therefore, only one-half mole of molecular nitrogen is required for each mole of ammonia.
What element is needed to create ammonia?
Nitrogen and Hydrogen.
How many moles of nitrogen and hydrogen are needed to get 10 moles of ammonia?
The reaction of nitrogen with hydrogen to form ammonia is: N2 +3H2 = 2NH3 Therefore to make 10 moles of ammonia you need 5 moles N2 and 15 moles H2
How many hydrogen molecules are needed to produce 525 grams of ammonia?
To produce 525 grams of ammonia (NH3), you would need 25 moles of ammonia. Since the balanced chemical equation for the reaction between hydrogen and nitrogen to form ammonia is 3H2 + N2 -> 2NH3, you would need 75 moles of hydrogen molecules (H2) to produce 525 grams of ammonia. This is equivalent to 4,500 molecules of hydrogen.
Why pure nitrogen and pure hydrogen are used in haber's process?
Pure nitrogen and pure hydrogen are used in Haber's process because they are the starting materials needed to produce ammonia. Nitrogen is the main component in the atmosphere, and hydrogen is readily available through various industrial processes. By using pure nitrogen and pure hydrogen, the reaction conditions can be controlled to optimize the production of ammonia.
How many moles of hydrogen are required to produce 18.00 moles of ammonia?
N2 + 3H2 -> 2NH3 The stoichiometric equation (or balanced equation) for the formation of ammonia from this we can read off the mole ratio between hydrogen and ammonia; 3M H2 needed to produce 2M NH3 times each by 9 (so the ratio remains the same and 18M NH3 is formed) 27M H2 needed to produce 18M NH3
How many moles of ammonia are needed to produce 3.5 mole of nitrogen?
That amount of ammonia contains two moles of hydrogen gas. One mole of hydrogen gas weighs 2.016 grams. Therfore 3.75 grams of ammonia contains two moles of hydrogen.
The reaction of hydrogen with nitrogen to produce ammonia is shown below. If there are 3 moles of nitrogen how many moles of hydrogen are needed to react completely with the nitrogen?
The balanced equation for the reaction is: 3H2 + N2 -> 2NH3 From the balanced equation, we can see that 3 moles of hydrogen are needed to react completely with 1 mole of nitrogen. So if there are 3 moles of nitrogen, you would need 9 moles of hydrogen to react completely.
How many grams of nitrogen are needed to produce 660 grams of ammonia?
The equation for the formation of ammonia from nitrogen and hydrogen gases at standard temperature and pressure is N2 + 3 H2 -> 2 NH3. The gram molecular mass of ammonia is 17.03 and the gram molecular mass of divalent nitrogen is 28.0134. Therefore, designating the unknown mass of nitrogen needed as p, the proportion 660/p equals (2 X 17.03)/28.0134 is valid. This expression is algebraically equivalent to 660 = (34.06)p/28.0134, for which p = 543 grams, rounded to the justified number of significant digits, limited by the three significant digits specified by the given number 600.
What mass of nitrogen is needed to make 34 grams of ammonia?
To find the mass of nitrogen needed to make ammonia, first determine the molar mass of ammonia (NH3) which is 17 g/mol. Since there is one nitrogen atom in ammonia, the nitrogen mass is 14 g/mol. To make 34 grams of ammonia, you would need 14 grams of nitrogen.
In order to get 1 mole of ammonia how many moles of nitrogen is needed?
Each mole of ammonia requires one mole of nitrogen atoms. However, the nitrogen in the air occurs as diatomic molecules; therefore, only one-half mole of molecular nitrogen is required for each mole of ammonia.
How many hydrogen atoms would be needed to bond with one nitrogen atom to stabilize all atoms?
Three hydrogen atoms would be needed to bond with one nitrogen atom and the name of this molecule is ammonia.
When 28 g of nitrogen and 6 g of hydrogen react 34 g of ammonia are produced. If 80 g of nitrogen react with 4 g of hydrogen how much ammonia will be produced?
Using the law of multiple proportions, we can see that the ratio of nitrogen to hydrogen in ammonia is 28:6 = 4.67:1. Therefore, for 80 g of nitrogen, 80/4.67 = 17.12 g of hydrogen would be needed to react completely. Since only 4 g of hydrogen is available, the limiting reactant is hydrogen and only 6 g of ammonia will be produced.
