What is the greatest amount of ammonia (NH3) in moles that can be made with 4.7 mol N2 and 6.9 mol H2? Which reactant is limiting? How many moles of excess reactant will be left after the reaction?

Question

Bentley Hood

Chemistry

8 July, 19:53

What is the greatest amount of ammonia (NH3) in moles that can be made with 4.7 mol N2 and 6.9 mol H2? Which reactant is limiting? How many moles of excess reactant will be left after the reaction?

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Answers (1)

Know the Answer?

Demarion Kirk · Answer 1

A. H2 is the limiting reactant.

B. The greatest mole of NH3 produced is 4.6 moles

C. The excess reactant is N2 and the leftover mole is 2.4 moles

Explanation:

The balanced equation for the reaction is given below:

N2 + 3H2 - > 2NH3

To obtain the greatest amount of NH3 that can be formed from the reaction of 4.7 mol of N2 and 6.9 mol of H2, we must first determine the limiting reactant.

A. Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 4.7 moles of N2 will react with = 4.7 x 3 = 14.1 moles of H2.

We can see from the calculations made above, that it will take a higher amount i. e 14.1 moles of H2 than 6.9 moles given to react with 4.7 moles of N2.

Therefore, H2 is the limiting reactant and N2 is the excess reactant.

B. Determination of the greatest amount of NH3 from the reaction.

The limiting reactant is always used to obtain the maximum yield of any reaction as all of it is used in the reaction. The limiting reactant for the above reaction is H2 and the greatest amount of NH3 is obtained as follow:

From the balanced equation above,

3 moles of H2 reacted to produce 2 moles of NH3.

Therefore, 6.9 moles of H2 will react to produce = (6.9 x 2) / 3 = 4.6 moles of NH3.

The greatest amount of NH3 produced is 4.6 moles

C. Determination of the moles of excess reactant that is leftover after the reaction.

This is illustrated below:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, Xmol of N2 will react with 6.9 moles of H2 i. e

Xmol of N2 = 6.9/3

Xmol of N2 = 2.3 moles

Therefore, 2.3 moles N2 reacted

The leftover mole of N2 is obtained as follow:

Mole of N2 given = 4.7 moles

Mole of N2 that react = 2.3 moles

Leftover mole of N2 =.?

Leftover mole of N2 = (Mole of N2 given) - (Mole of N2 that react)

Leftover mole of N2 = 4.7 - 2.3

Leftover mole of N2 = 2.4 moles