Balanced equation always and always first!
N2 + 3H2 >> 2NH3
As you see the dimensional analysis shows three mols H2 to one mol N2. ( mols are always coefficients )
So, three times two is 6.
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
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
28 grams of Nitrogen is necessary to produce 34 grams of ammonia.
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.
3H2 + N2 <------> 2NH3 is the balanced equation for Hydrogen and Nitrogen making ammonia. 3 moles of H2 produces two moles of ammonia and thus to make 6 moles requires 9 moles of Hydrogen.
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
Nitrogen and Hydrogen.
21 L
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
To form ammonia, balanced reaction is N(2) + 3H(2) ---> 2NH(3) + H(2)O. As you can see for 1 mole of nitrogen three moles of hydrogen is required. Hence for your question, 3 moles nitrogen is required to satisfy the ratio.
Three hydrogen atoms would be needed to bond with one nitrogen atom and the name of this molecule is ammonia.
According balanced equation ,3 mols are needed. So 3mol shoul be mixed
28 grams of Nitrogen is necessary to produce 34 grams of ammonia.
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.
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.
N2 + 3H2 -> 2NH3 3 moles hydrogen gas. You should know that because of the formula of ammonia.
Nitrogen is needed for all life forms, including plants. Plants absorb nitrogen from soil through their roots in the form of nitrate and ammonia.