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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.
What is the mass percentage of nitrogen in ammonia?
The mass percentage of nitrogen in ammonia (NH3) is 82.35%. This is calculated by dividing the mass of nitrogen in one mole of ammonia by the molar mass of ammonia, and then multiplying by 100 to get the percentage.
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.
How many grams of oxygen are needed to react with 6.78 grams of ammonia?
To calculate the number of grams of oxygen needed to react with 6.78 grams of ammonia, we first write out the balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO) and water (H2O). Then we use the stoichiometry of the equation to find the molar ratio between ammonia and oxygen. Finally, we convert the mass of ammonia to moles and then use the molar ratio to find the mass of oxygen needed.
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
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.
What is the mass percentage of nitrogen in ammonia?
The mass percentage of nitrogen in ammonia (NH3) is 82.35%. This is calculated by dividing the mass of nitrogen in one mole of ammonia by the molar mass of ammonia, and then multiplying by 100 to get the percentage.
What is the mass of nitrogen in 125 g of ammonia?
In ammonia (NH3), the molar mass is 17 g/mol. To find the mass of nitrogen in 125 g of ammonia, first, calculate the number of moles of ammonia in 125 g. Then, multiply the moles of ammonia by the molar ratio of nitrogen in ammonia (1 mol of N for every 1 mol of NH3), and finally, multiply by the molar mass of nitrogen (14 g/mol) to find the mass of nitrogen. This will give the mass of nitrogen in 125 g of ammonia.
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 ammonia is formed when 3.80 g of nitrogen gas react with hydrogen?
To calculate the mass of ammonia formed, first write out the balanced chemical equation for the reaction between nitrogen and hydrogen to form ammonia: N₂ + 3H₂ → 2NH₃ Next, calculate the moles of nitrogen in 3.80 g using the molar mass of nitrogen (N₂). Then use the mole ratio from the balanced equation to determine the moles of ammonia formed. Finally, convert the moles of ammonia to grams using the molar mass of ammonia (NH₃) to find the mass formed.
Percent age of nitrogen in urea and ammonia?
it can be calculated using the formula percentage composition of N =Gram molecular mass of nitrogen in the compound/ Gram molecular mass of compound *100
How many grams of oxygen are needed to react with 6.78 grams of ammonia?
To calculate the number of grams of oxygen needed to react with 6.78 grams of ammonia, we first write out the balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO) and water (H2O). Then we use the stoichiometry of the equation to find the molar ratio between ammonia and oxygen. Finally, we convert the mass of ammonia to moles and then use the molar ratio to find the mass of oxygen needed.
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
If 250.3 L of hydrogen gas (d 0.0899gL ) reacts with an excess of nitrogen gas what mass of ammonia would be produced?
To determine the mass of ammonia produced, you first need to calculate the moles of hydrogen gas present. Then, you can use the stoichiometry of the balanced chemical equation for the reaction between hydrogen and nitrogen to find the moles of ammonia produced. Finally, using the molar mass of ammonia, you can convert moles to grams to find the mass of ammonia produced.
How much nitrogen is in 7.5 grams of ammonia?
First you need to find the atomic masses of each element involved in the compound NH3, and add them up to find the total molecular mass of ammonia.Nitrogen = 14.0 gramsHydrogen = 1.01 × 3 atoms = 3.03 grams----------------------------------------------------Ammonia = 17.03 gramsThen you take the mass of nitrogen in one molecule and divide it by the total mass to find the percent composition.14.0 grams Nitrogen ÷ 17.03 grams Ammonia = .822 = 82.2% nitrogen in ammoniaThen you simply need to take 82.2% of 7.5 grams to find how much nitrogen is in that particular amount.82.2% × 7.50 = 6.17 grams of nitrogen in 7.50 grams of ammonia
What is atomic mass of ammonia?
The atomic mass of ammonia (NH3) is the sum of the atomic masses of its constituent atoms. Nitrogen has an atomic mass of approximately 14.01 amu, while hydrogen has an atomic mass of approximately 1.01 amu. Therefore, the atomic mass of ammonia is approximately 17.03 amu.
How many tons does ammonia make?
The production of ammonia can vary depending on the method used. However, typically, one ton of ammonia is produced from around 1.8 to 2 tons of nitrogen gas and three to five tons of hydrogen gas through the Haber-Bosch process.
