Nitrogen N2 gas and hydrogen H2 gas react to form ammonia NH3 gas. Suppose you have 11.0 mol of N2 and 1.0 mol of H2 in a reactor. Calculate the largest amount of NH3 that could be produced. Round your answer to the nearest 0.1 mol.

362.7 g of NH3

Explanation:

The balanced equation of the given reaction is represented as;

N2 (g) + 3H2 (g) -> 2NH3 (g)

From observation, we can see that the mole ratio between the elements in the reaction which are: N2, H2 and NH3 have a mole ratio 1 : 3 : 2

As we can see that N2 : H2 = 1:3

It means that the reaction will consume three times more moles of hydrogen than the moles of nitrogen consumed. Thus, the H2(hydrogen gas) will limit the amount of N2 (Nitrogen gas) that will take part in the reaction.

Thus,

No. of moles of N2 that will take part in the reaction is;

1 mole of H2 × 1 mole of N2/3 moles of H2

The moles of H2 will cancel out to give;

1/3 moles of N2

We are told we have 11.0 mol of N2.

Thus, remaining N2 will be; 11 - ⅓ = 32/3 moles of N2

The mole ration between N2 and NH3 as a earlier seen is 1:2.

Thus,

We have;

32/3 moles of N2 × 2 moles of NH3/1 mole of N2

The mole of N2 will cancel out to give;

64/3 moles of NH3

Now, atomic mass of NH3 = 17 g/mol

Thus, max amount of NH3 that could be produced is;

64/3 × 17 = 362.7 g

b) mutualism

hope this : )

i'm assuming that the correct answer is indeed the first one/ "a"..