Mass(g)of N = 35.8 g NH4NO3 X 28 g of N / 80 g NH4NO3 = 12.53 g N
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The volume occupied by 3 moles of nitrogen gas will be different, depending on the temperature and pressure of the gas.
Ammonium Nitrate = NH4NO3 or N2H4O3 the total mass of a ammonium nitrate molecule is as follows: 2*14u + 4*1u + 3*16u = 28u + 4u + 48u = 80u now, the total mass of Nitrogen in one ammonium nitrate is 2*14u = 28u. Then, we divide 28 (the mass of Nitrogen) by 80 (total mass)= 28/80 = 0.35, which is the ratio for Nitrogen mass divided by total mass. then, we get the 48.5 grams (total mass) and multiply it by this ratio (0.35): 48.5 * 0.35 = 16.975 grams of Nitrogen in 48.5 grams of Ammonium Nitrate.
9 atoms Here is the formula for ammonium nitrate. NH4NO3 This shows there are two nitrogen atoms, 4 hydrogen, and 3 oxygen. So 1 (nitrogen) + 1(nitrogen) + 4(hydrogen) + 3(oxygen) = 9 atoms
1 mole of Ammonium Nitrate= 80g (R.A.M of compound) Therefore 8g of Ammonium nitrate= 8g/80g= 0.1 mole (moles = mass given over molar mass)
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The chemical formula for ammonium nitrate is NH4NO3. It consists of 3 elements: nitrogen, hydrogen and oxygen.
The volume occupied by 3 moles of nitrogen gas will be different, depending on the temperature and pressure of the gas.
Ammonium Nitrate = NH4NO3 or N2H4O3 the total mass of a ammonium nitrate molecule is as follows: 2*14u + 4*1u + 3*16u = 28u + 4u + 48u = 80u now, the total mass of Nitrogen in one ammonium nitrate is 2*14u = 28u. Then, we divide 28 (the mass of Nitrogen) by 80 (total mass)= 28/80 = 0.35, which is the ratio for Nitrogen mass divided by total mass. then, we get the 48.5 grams (total mass) and multiply it by this ratio (0.35): 48.5 * 0.35 = 16.975 grams of Nitrogen in 48.5 grams of Ammonium Nitrate.
1 mole of ammonium nitrate produces one mole of nitrogen. Actually the amount (in moles) of nitrogen will depend on how much NH4NO3 you are starting with, what other reactant you are combining it with and whether or not the NH4NO3 completely reacts. Since you will never be able to retrieve all of the nitrogen (either the NH4 or the NO3 will retain some nitrogen depending upon the reaction), you can reasonably expect to get 1 mole of N2 for each 14.01 grams of Ammonium nitrate that COMPLETELY reacts.
9 atoms Here is the formula for ammonium nitrate. NH4NO3 This shows there are two nitrogen atoms, 4 hydrogen, and 3 oxygen. So 1 (nitrogen) + 1(nitrogen) + 4(hydrogen) + 3(oxygen) = 9 atoms
Ammonium nitrate(NH4NO3) composition:Nitrogen = 2 atomsOxygen = 3 atomsHydrogen = 4 atoms
Urea is 46-0-0, and ammonium nitrate is 34-0-0. Both of these products contain nothing but nitrogen, but the nitrogen is at different percentages. Urea contains 920 actual pounds of nitrogen per ton, and ammonium nitrate contains 680 actual pounds of nitrogen by ton (2,000 x 46% = 920 and 2,000 x 34% = 680)
Nine atoms represent the empirical formula of ammonium nitrate. Namely, 2 nitrogen atoms, 3 oxygen atoms and 4 hydrogen atoms.
1 mole of Ammonium Nitrate= 80g (R.A.M of compound) Therefore 8g of Ammonium nitrate= 8g/80g= 0.1 mole (moles = mass given over molar mass)
2
0.0999458293608 moles