How many moles of NH3 are produced if 2.5 grams of N2 reacts?

How many moles of NH3 are produced if 2.5 grams of N2 reacts?

According to the equation in the question, For every mole of N2 needed, 3 moles of H2 are needed and 2 moles of NH3 are produced. For every 2.5 moles of N2 needed, 7.5 moles of H2 are needed and 5.0 moles of NH3 are produced.

How many moles of NH3 will be produced if 12 moles H2 react completely with N2?

8 moles
Then, you can calculate how many moles of ammonia are produced when 12 moles of hydrogen reacts. Keep in mind, you also need a certain amount of nitrogen gas, but that’s not what the question is asking, so we can ignore it. Therefore, 12 moles of hydrogen makes 8 moles of ammonia.

READ ALSO:   What happens when you click No thanks on LinkedIn?

How many moles of N2 reacted if 0.80 mole NH3 is produced?

Ammonia is produced by the reaction of nitrogen and hydrogen. How many moles of N2 reacted if 0.80 mol of NH3 is produced? Express your answer to two significant figures and include the appropriate units. Answer: 0.40 moles of N2.

How many moles of water can be produced from the reaction of 28g c3h 8?

2.5⋅mol .

How many moles of NH3 will be formed if 18 moles of H2 is used?

How many moles of NH3 are produced when 18 mol of H2 are reacted? Solution According to the balanced equation, 2 moles of NH3 are produced when 3 moles of H2 react.

What is the mole ratio of NH3 N2?

The chemical equation of the production of ammonia is shown in figure 1 and it shows that the mole ratio for ammonia and nitrogen gas is 2:1.

How much ammonia NH3 is produced from two moles of nitrogen gas?

READ ALSO:   How do you write secret codes?

The balanced equation shows that one mole of nitrogen (N2) will produce two moles of ammonia. 73 grams of ammonia is (7317)⋅(gramsNH3gramsmolNH3)=4.29 moles ammonia.

How many grams NH3 can be produced?

Since the molar ratio of N2 to Nh3 is 2:1, you can make twice as much NH3 by using N2. Therefore, you can make a maximum of 4.5 mol of NH3 assuming 100\% yield.