3,44 moles H2 react with 1,146 moles NH3. The limiting reactant is hydrogen. O,244 moles N2 remain. 19,5 g NH3 are obtained.
1 g of ammonia (NH3) is equal to 0,059 mol.
The mass of 3 mol of ammonia is 51,093 g; the number of ammonia molecules in 3 moles is18,066422571.10e23.
Take a multiple of three moles of diborane plus a multiple of six moles of ammonia, mix them together, heat them to 300 degrees Celsius and stand by with a fire extinguisher because reacting three moles of diborane with six moles of ammonia liberates 12 moles of hydrogen.
N2 + 3H2 -> 2NH3 3 moles hydrogen gas. You should know that because of the formula of ammonia.
How many moles of ammonia is 170000grams?
N2 + 3H2 -----> 2NH3 so 3 moles of hydrogen produce 2 moles of ammonia. Therefore 12.0 moles of hydrogen will produce 8 moles of ammonia.
0,522 moles of ammonia contain 3,143.10e23 molecules of NH3.
8,038 moles of ammonia were produced.
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
2N + 3H2 -> 2NH3 18 moles H2 (2 moles NH3/3 moles H2) = 12 mole ammonia
4NH3 + 3O2 -----> 2N2 + 6H2O 4 moles of ammonia react with 3 of oxygen. So 10 moles of ammonia requires 7.5 moles of oxygen.
3 moles of ammonia is 51grams. One mole is 17 grams.
6 moles COULD be produced
N2 + 3H2 ---> 2NH3 so 3 moles of Hydrogen produces 2 moles of ammonia. Thus 1.8 moles will produce 1.8/3 x 2 = 1.2 moles of ammonia.
The molar mass of ammonia is about 17 grams, so that 3 moles would have a mass of 51 grams.
3,44 moles H2 react with 1,146 moles NH3. The limiting reactant is hydrogen. O,244 moles N2 remain. 19,5 g NH3 are obtained.