amonia, NH3, is produced from the reaction of nitrogen and hydrogen gases?

Question

in one experiment 2.04g of H2 is mixed with 11.35g of N2. what is the theoretical yield of NH3? I.E. how many grams of NH3 chould be formed?

Answer 1

If the reaction would "go" it is written: 3H2 + N2 = 2 (NH3)
Fill in the atomic masses of H = 1.00797 and N = 14.0067:
[3 x 2 x (1.00797)] +[ 2 x (14.0067)] = 2 x (14.0067) x 3 x (1.00797) = 34.06122
This is the stochiometric ratio of elements needed. You have 2.04 gm of H2, so multiply each term by 2.04 / 6.04782 to get:
2.04 + 9.4492 = 11.4892. In other words, all of the H2 combines with about 9.45 gm of N2 to yield about 11.49 gm of ammonia with about 1.9 gm of N2 left over.

Answer 2

the reaction might study: 3 H2 + N2 = 2 NH3 First you're able to be sure the proscribing component (do you have extra N2 or extra H2, yet do no longer forget that for each mole of N2 your prefer 3 moles of H2). H2 has a molecular weight of two.002 and N2 has a molecular weight of 14.02. Dividing the grams of each substance via it particularly is molecular weight yields the # moles of that substance. There are a million.19 moles of H2 and .945 moles of N2. to apply all of the N2 you will possibly wish three times that quantity in H2 (2.80 4 moles) and you purely have a million.19 moles. consequently H2 is your proscribing agent. for each 3 moles of H2 you get 2 moles of NH3. so which you're taking the form of moles of H2 and divide via 3 and multiply via 2. You get .seventy 9 moles of NH3. Multiplying this via its molecular weight (17.02) delivers a yield of 13.40 seven grams. wish this enables!

Answer 3

Use stoichiometry.