56.50g of N2 and 48.65g of H2. Ends up collecting 60.05g of NH3. How many grams of NH3 should have theoretically be produced from the reaction? percent yield? How much of the excess reagent is left ?

Question

show work please, i need major help lol

Answer 1

If N₂ reacts with H₂ to form NH₃, the balanced chemical equation would be

N₂ + 3H₂ → 2NH₃

This question is asking you to find the limiting reactant-which reactant there is less of. This does not always mean the smaller mass, but the smaller amount of product that each would produce.

First step is to convert both masses into moles. Multiply the mass given by each molar mass. The molar mass of N₂ is 28.00 g/mol. The molar mass of H₂ is 2.02 g/mol.

The next step is to convert from the moles of each reactant to moles of the product. Whichever reactant yields a smaller number of moles of NH₃ is the limiting reactant. This is how much NH₃ can have been theoretically produced.

56.50 g N2	1 mol N2	2 mol NH3
	28.00 g N2	1 mol N2

There would be 4.04 mol NH₃ produced from 56.50 g N₂.

48.65 g H2	1 mol H2	2 mol NH3
	2.02 g H2	3 mol H2

There would be 16.06 mol NH₃ produced from 48.65 g H₂.

Because there are fewer moles produced from N₂, it is the limiting reactant, and it will determine how much NH₃ can be produced. Use the 4.04 mol NH₃ and multiply by its molar mass of 17.03 g/mol:

4.04 mol NH3	17.03 g NH3
	1 mol NH3

68.80 g of NH₃ is produced.

Not all the H₂ was used up, so this is the excess reagent. To find the amount of H₂ used up in the reaction, start with the mass of the limiting reagent, mass of N₂, and convert it to grams of H₂. Multiply across.

56.50 g N2	1 mol N2	3 mol H2	2.02 g H2
	28.00 g N2	1 mol N2	1 mol H2

The mass of H₂ used up was 12.23 g.

The amount of H₂ remaining was therefore 48.65g – 12.23g = 36.42g H₂.