How many grams of hydrogen are necessary to react completely with 50.0 g of nitrogen in the above reaction? How many grams of ammonia are produced in the reaction
First, we need to write and balance the chemical equation for the reaction:
N2 + 3H2 → 2NH3
From the balanced equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
1 mole of N2 has a molar mass of 28.02 g
So, 50.0 g of N2 is equal to 50.0 / 28.02 ≈ 1.784 moles of N2
To react completely with the nitrogen, we need 3 moles of hydrogen for every mole of nitrogen, so we need:
1.784 moles of N2 * 3 moles H2/mole N2 = 5.35 moles H2
The molar mass of hydrogen is 2.016 g/mol, so the mass of hydrogen needed is:
5.35 moles of H2 * 2.016 g/mol ≈ 10.79 g of H2
Therefore, 10.79 grams of hydrogen are necessary to react completely with 50.0 g of nitrogen.
To find the mass of ammonia produced, we first calculate the number of moles of ammonia produced:
From the balanced equation, 1 mole of N2 produces 2 moles of NH3
So, from 1.784 moles of N2, we get 1.784 * 2 = 3.568 moles of NH3
The molar mass of ammonia (NH3) is 17.03 g/mol, so the mass of ammonia produced is:
3.568 moles of NH3 * 17.03 g/mol ≈ 60.77 g of NH3
Therefore, approximately 60.77 grams of ammonia are produced in the reaction.