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Your responses to a subset of these questions are to be submitted using the HELP! link on the Problem Set Page on your Personal Page. Given the reaction Initially (before any reaction occurs) a 1.00 liter reaction vessel al 400 degree C contains 0.53 moles of 02(g) and 0.953 moles of NH3(g) and no water or nitrogen. Consider the following: a) If 0.064 moles of 02(g) react, how many moles of NH_3(g) must react and how many moles of H_2O(g) and N2(g) are formed? How many moles of O_2(g). NH_3(g). H_2O(g) and N_2(g) remain after completion of the reaction?

Explanation / Answer

SOLUTION:

3O2 + 4NH3 <----> 6H2O + 2N2

Frombalanced chemical equatio we know 3 moles of O2 reat with 4 moles of NH3 to give 6 moles of H2O and 2 moles of N2.

(A) moles of NH3 that will react with 0.064 moles of O2:

3 moles of O2 react with 4 mols of NH3

Or 3 Moles of O2 = 4 moles of NH3

Therefore 1 mole of O2 will react with = 4/3 moles of NH3

Hence 0.064 moles of O2 will react with (4/3 ) X 0.064 = 0.085 moles of NH3

(B) Moles of H2O formed

3 moles of O2 form 6 moles of water,

3 moles of O2 = 6 moles of H2O

1 moles of O2 will form 6/3 = 2 moles of water

therefore 0.064 moles of O2 will form 2 X 0.064 = 0.128 moles of Water

(C) Moles of N2 = (2 / 3) X 0.064 = 0.043 moles (using above methodology)

(D) Moles of O2 after reaction = initial moles - moles reacted = 0.53 - 0.064 = 0.466 moles

Moles of NH3 that remain afer the reaction = 0.953 - 0.085 = 0.868 moles.

Moles of Water formed = 0.128 moles

Moles f N2 formed = 0.043 moles.

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