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How many mols are in 400g MgO?

The answer is 9,92 moles.


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 are there in 400g of calcium cabonate?

400 g of calcium cabonate is equal to 3,996 moles.


How many moles of NH3 is in 23g of NH3?

Atomic mass of NH3 is 17.So there are 1.353mol.


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


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.