Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:

N2 (g) + 3H2 (g) ? 2NH3 (g)

1. What is the maximum mass of ammonia that can be produced from a mixture of 163.3 g of N2 and 38.77 g of H2? g

2. Which element would be completely consumed? (enter nitrogen or hydrogen)

3. What mass of the starting material would remain unreacted?

Look at how many grams of ammonia the N2 would produce and how many grams of ammonia the H2 would produce.

Watch the mole ratios!

Whichever produces the smallest amount of ammonia is the limiting reactant.

HINT: After you have found which starting material was used up completely, you know that some of the

other starting material is still unreacted. Subtract the reacted material (which you must calculate) from the

original amount you had in the beginning, thus finding how much remains unreacted

Answer: 1) Maximum mass of ammonia 198.57g

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g

Explanation:

In order to calculate the Mass of ammonia, we first check the Equation is actually Balance:

N2 (g) + 3H2 (g) ⟶2NH3 (g)

Both equal amount of atoms side to side.

Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. (Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)

Moles of ammonia produced with 163.3g N2 (g) ⟶ 163.3g N2 (g) x (1mol N2 (g) / 28.01 g N2 (g)) x (2 mol NH3 (g) / 1 mol N2 (g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x (1mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) = 12.79 mol NH3

As we can see the amount of NH3 formed with the N2 is the lowest one, therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it. We proceed calculating the maximum mass of NH3 from the 163.3g of N2.

11.66 mol NH3 x (17.03 g NH3 / 1mol NH3) = 198.57 g NH3

In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:

163.3g N2 x (1mol N2/28.01 g N2) x (3 mol H2 / 1 mol N2) x (2.02 g H2 / 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left