Balanced equation. N2 + 3H2 --> 2NH3 1.4 moles H2 (2 moles NH3/3 moles H2) = 0.93 moles NH3 produced =======================
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.
2.49x10-1mol NH3Source: e2020
To find the grams of F2 required for the reaction, first calculate moles of NH3: 69.3 g NH3 / 17.03 g/mol = 4.07 moles NH3. From the balanced equation, 5 moles of NH3 react with 2 moles of F2, so you need 4.07 moles NH3 * (2 moles F2 / 5 moles NH3) * 38.0 g/mol = 30.6 g of F2 for complete reaction.
Using the molar mass of nh3, we find that we have 2.5 moles of nh3. Since 3 moles of h2o are produced per 2 moles of nh3, we see that we will produce 3.75 moles of h2o. This is equivalent to around 3.79 g.
The answer is 9,92 moles.
0,044 moles of NH3 can be produced.
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?
Balanced equation. N2 + 3H2 --> 2NH3 1.4 moles H2 (2 moles NH3/3 moles H2) = 0.93 moles NH3 produced =======================
400 g of calcium cabonate is equal to 3,996 moles.
Atomic mass of NH3 is 17.So there are 1.353mol.
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.
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.
2.49x10-1mol NH3Source: e2020
To find the grams of F2 required for the reaction, first calculate moles of NH3: 69.3 g NH3 / 17.03 g/mol = 4.07 moles NH3. From the balanced equation, 5 moles of NH3 react with 2 moles of F2, so you need 4.07 moles NH3 * (2 moles F2 / 5 moles NH3) * 38.0 g/mol = 30.6 g of F2 for complete reaction.