A) write the word equation for the reaction of barium nitride (ba3n2) with potassium.

b) write the balanced formula equation for this reaction.

c) if 29.0 g of potassium reacts with 66.5 g of barium nitride, what mass of potassium nitride will be produced?

d) identify which reactant is the limiting reactant and which is the excess reactant.

e) how many grams of the excess reactant reacts and how many grams are left over?

Explanation:

In order to balance it, we need to have the same number of atoms of each element on both sides of the equation. There are two atoms of nitrogen on the left, so we need to put 2 in front of K₃N. Now, we have six atoms of potassium on the right, so we need to put 6 in front of K on the left. Finally, there are three atoms of barium on the left, so we put 3 before Ba on the tight. Which means:

Ba₃N₂ + 6K = 2K₃N + 3Ba

Now, we can do the work. First, we determine the molar mass of each reactant ( from the periodic table). Molar mass of the barium is 137, potassium 39 and nitrogen 14.

Ba₃N₂₂ has molar mass of 3Ba and 2N, which means 3 • 137 + 2 • 14 = 439. That means that one mole of Ba₃N₂ weights 439 grams.

We are given grams of reactants, but in order to find the limiting and the excess reactant, we need to transfer it into moles.

We are given 66.5 grams of Ba₃N₂ and we know that 439 grams equals 1 mole. We want to know how many moles there are in 66.5 grams, so the answer is 66.5 / 439 = 0.15 moles.

Let's do the same for potassium. We are given 29 grams of K and we know that 1 mole has 39 grams. We want to know how many moles of K are there in 29 grams, so the answer is 29 / 39 = 0.74 moles.

We now know that 0.15 moles of Ba₃N₂ reacted with 0.74 moles of K. From the balanced equation we see that 1 mole of Ba₃N₂ reacts with 6 moles of K, so the ratio has to be 1:6.

Now let's find limiting and excess reactant. That means that in our reaction, there are more (or less) of one reactant then needed.

We know that we had 0.15 moles of Ba₃N₂ reacting. Let's pretend we don't know the moles of K and let's see with which amount of K should 0.15 moles of barium nitride react, if the ratio is 1:6.

0.15 moles of Ba₃N₂ : x moles of K = 1:6

x = 0.9 moles of K

So, for the completed reaction we need to have 0.9 moles of K, but we previously calculated that we had 0.74. That means that there is less K then needed, so potassium is our limiting reactant, which obviously means that Ba₃N₂ is our excess reactant.

Now, we need to find how many moles of Ba₃N₂ there needed to be for a completed reaction

x moles of Ba₃N₂ : 0.74 moles of K = 1:6

x = 0.124 moles of Ba₃N₂

So we needed to have 0.124 moles, but we had 0.15 of Ba₃N₂, which is 0.15 - 0.124 = 0.026 moles in excess.

If we want to find how many grams that is, we only multiply it with molar mass of Ba₃N₂:

0.026 • 439 = 11.4 grams

That means that only 66.5 - 11.4 = 55.1 grams of Ba₃N₂ reacted.

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