One of the steps m the commercial process for converting ammonia to nitric acid

Question

One of the steps m the commercial process for converting ammonia to nitric acid is the conversion of NH_3 to NO 4NH_3 (g) + 5O_2(g) rightarrow 4 NO(g) + 6 H_2 O(g) In a certain experiment. 2 10 g of NH3 reacts with 3 85 g of O_2 Which is the limiting reactant? How many grams of NO and of H_2O form? Enter your answers numerically separated by a comma. How many grams of the excess reactant remain after the limiting reactant is completely consumed? Express your answer in grams to three significant figures.

Explanation / Answer

B)

mol of O2 = mass/MW = 3.85/32 = 0.1203125 mol of O2 will react

5 mol of O" --< 4 mol of NO

so

0.1203125 mol of O2 --> 4/5*0.1203125 = 0.09625 mol of NO

mass of NO = mol*MW = 0.09625*30 = 0.09625 g of NO

now, for H2O:

5 mol of O2 = 6 mol of H2O

so..

0.1203125 mol --> 6/5*0.1203125 = 0.144375 mol of H2O

mass = mol*MW = 0.144375*18 = 2.59875 g

C)

excess reatant:

mol of NH3 = mass/MW = 2.10/17 = 0.12352 mol of NH3

we know 0.1203125 mol of O2 reacts

so ratio is 4/5--- > 4/5*0.1203125 = 0.09625 mol of NH3 reacts

NH3 left = 0.12352-0.09625 = 0.02727 mol left

mass = mol*Mw = 0.02727*17 = 0.46359 g of NH3 --> 3 sig ifg -- > 0.464 g

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