How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
0,044 moles of NH3 can be produced.
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
The balanced chemical equation for the reaction is: 3H2 + N2 -> 2NH3 From the stoichiometry of the balanced equation, 3 moles of H2 produces 2 moles of NH3. Therefore, if you have 1.5 moles of H2, you will produce 1.5 moles * (2 moles NH3 / 3 moles H2) = 1 mole of NH3.
The answer is 1,57.10e27 molecules.
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
The balanced chemical equation for the reaction between H2 and NH3 is: 3H2 + N2 → 2NH3 From the equation, we can see that 3 moles of H2 produce 2 moles of NH3. Therefore, when 1.2 moles of H2 react, we can calculate the moles of NH3 produced as: 1.2 mol H2 * (2 mol NH3 / 3 mol H2) = 0.8 mol NH3.
Balanced equation. N2 + 3H2 --> 2NH3 1.4 moles H2 (2 moles NH3/3 moles H2) = 0.93 moles NH3 produced =======================
0,044 moles of NH3 can be produced.
The balanced chemical equation for the reaction between H2 and NH3 is: 3H2 + N2 -> 2NH3 From the balanced equation, we can see that 3 moles of H2 produce 2 moles of NH3. Therefore, 1.8 moles of H2 will produce 1.2 moles of NH3.
0,3 moles of nitrogen reacted.
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
The balanced chemical equation for the reaction is: 3H2 + N2 -> 2NH3 From the stoichiometry of the balanced equation, 3 moles of H2 produces 2 moles of NH3. Therefore, if you have 1.5 moles of H2, you will produce 1.5 moles * (2 moles NH3 / 3 moles H2) = 1 mole of NH3.
The reaction would be H2 + 3N2 ==>2NH3moles H2 used = 5.69104 g x 1 mole/2.00 = 2.84552 moles H2moles NH3 produced (assuming N2 is NOT limiting) = 2 moles NH3/mole H2 x 2.84552 moles H2 = 5.69104 moles NH3 producedMolecules of NH3 produced = 5.69104 moles x 6.02x10^23 molecules/mole = 3.4x10^24 molecules
The answer is 1,57.10e27 molecules.
8 mol
Balanced equation is N2 + 3H2 ==> 2NH33.07104 g H2 x 1 mol/1.0079 g = 1.7679 moles H2 presentmoles NH3 produced = 2/3 x 1.7679 moles = 1.1786 moles NH3 formedmolecules NH3 = 1.1786 moles x 6.022x10^23 molecules/mole = 7.098x10^23 molecules (4 sig figs based on sig figs used in 6.022x10^23)