For the balanced equation: N2 + 3H2 = 2NH3. What is the mass of the excess reagent when 84.06g of N2 and 22.18g of H2 are combined

Answer 1

First, we need to convert each reactant's amount from grams to moles...

84.06 g N2 x 1 mol/28.02 g = 3.00 mol N2

22.18 g H2 x 1 mol H2/2.02 g = 10.98 mol H2

Then, we need to consider the balanced reaction for the reaction between N2 and H2...

1 N2 + 3 H2 --> 2 NH3

So, for every 3 moles of H2, we need 1 mole of N2. That's a 3:1 ratio of moles of H2 to moles of N2. Let's see where we are in relation to that. Let's divide the 10.98 mol H2 by the 3.00 moles of N2 to see if we have a ratio of 3:1 as dictated by the balanced reaction, or if the ratio is higher or lower than that.

10.98 mol H2

-----------------

3.00 mol N2

= 3.66 mol H2 to every one mole N2. That's more than the 3:1 moles of H2 to N2 required by the reaction. So, there are more moles of H2 than required by the reaction which makes H2 the excess reactant and N2 the limiting reactant.

Now, let's figure out how many moles and then grams of H2 are left over after the limiting reactant runs out. The limiting reactant is N2 and all of those moles will run out, so let's base our calculation from that. We will convert the moles of N2 which reacted to moles of H2 which reacted using the 3:1 ratio in the balanced reaction:

3.00 mol N2 x 3 molH2/1 mol N2 = 9.00 moles of H2 reacted.

Now let's subtract the 9.00 moles of H2 which reacted from the initial amount of 10.98 moles (way back when)...

10.98 moles H2 - 9.00 moles H2 = 1.98 moles left over.

The question asks for mass of excess reactant left over, so let's convert the 1.98 moles of H2 to grams using the formula weight for H2 of 2.02 g/mol...

1.98 moles H2 excess x 2.02 g H2/1 mol = 3.996 grams H2 in excess.

According to the rules for significant figures, we should probably express this in 4 digits as all of the introductory numbers were presented in 4 digits...

so, 3.996 grams H2 in excess should stand as our final answer.

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