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Atomic Mass of NH3 is 17.So there are 1.353mol.
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How many moles of NH3 can be produced from the reaction of 75g of N2?
0,044 moles of NH3 can be produced.
How many moles of NH3 are produced when 1.2 moles of H2 reacts?
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 moles H2 reacts?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
How many moles of NH3 are produced when 1.4 moles H2 reacts?
Balanced equation. N2 + 3H2 --> 2NH3 1.4 moles H2 (2 moles NH3/3 moles H2) = 0.93 moles NH3 produced =======================
How many moles of NH3 are produced when 1.2 mol of H2 reacts?
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 produce how many moles of N2 must have been used?
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
How many moles of nh3 are produced from 1.5 moles h2?
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.
How many moles is 1.50 X1023 molecules of NH3?
2.49x10-1mol NH3Source: e2020
How many moles of (NH4)2SO4 do you have?
0.758 moles of NH3 is the amount of moles in 50 grams of NH42SO4.
5f2 2nh3 yields n2f4 6hf if you have 69.3 g nh3 how many grams of f2 are required for complete reaction?
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.
How many moles of oxygen gas are needed to react with 10.0 mol of ammonia?
The balanced chemical equation for the reaction between ammonia (NH3) and oxygen gas (O2) is 4 NH3 + 5 O2 → 4 NO + 6 H2O. This means that 5 moles of O2 are needed to react with 4 moles of NH3. With 10.0 moles of NH3, you would need 12.5 moles of O2 (10.0 moles NH3 x 5 moles O2 / 4 moles NH3).
How many grams of oxygen are needed to react completely with 200.0 g of ammonia NH3?
The balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) is 4NH3 + 3O2 → 2N2 + 6H2O. From the equation, we can see that 3 moles of O2 are needed to react with 4 moles of NH3. This means the molar ratio of O2 to NH3 is 3:4. First, calculate the number of moles of NH3 in 200.0 g: 200.0 g NH3 / 17.03 g/mol NH3 = 11.75 moles NH3 Now, calculate the number of moles of O2 needed using the molar ratio: 11.75 moles NH3 * (3 moles O2 / 4 moles NH3) = 8.81 moles O2 Finally, convert moles of O2 to grams: 8.81 moles O2 * 32 g/mol O2 = 282.0 g O2.
