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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
What mass of nitrogen and hydrogen needed to produce 91.3 Kg of ammonia?
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
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
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.
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 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
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.
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.
What mass of nitrogen and hydrogen needed to produce 91.3 Kg of ammonia?
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
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.
If 6 liters of hydrogen gas are used how many liters of nitrogen gas will be needed for the above reaction at standard temperature and pressure?
Assuming you are referring to the reaction of hydrogen and nitrogen to form ammonia, the balanced equation is: 3H2 + N2 → 2NH3 From the equation, 3 liters of hydrogen gas react with 1 liter of nitrogen gas. Therefore, if 6 liters of hydrogen gas are used, you would need 2 liters of nitrogen gas.
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.
Gaseous ammonia is made by the following reaction n2 plus 3h 2nh3 if you start with 7 l of nitrogen gas how many liters of hydrogen are needed for a complete reaction?
According to the balanced chemical equation, for every 1 mole of nitrogen gas (N2), 3 moles of hydrogen gas (H2) are needed. Since the volume of a gas is directly proportional to the number of moles, you would need 21 liters of hydrogen gas (3 times 7 liters) to react completely with 7 liters of nitrogen gas to produce 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.
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
If 6 liters of hydrogen gas are used how many liters of nitrogen gas will be needed for the above reaction at STP?
For the reaction N₂ + 3H₂ → 2NH₃, the mole ratio of hydrogen gas to nitrogen gas is 3:1. Since 6 liters of hydrogen gas is used, you would need 2 liters of nitrogen gas at STP for this reaction according to the stoichiometry of the reaction.
