One of the steps in the commercial process for coveting ammonia to nitric acid i

Question

One of the steps in the commercial process for coveting ammonia to nitric acid is the conversion of NH_3 to NO: 4NH_3(g) + 5O_2(g) rightarrow 4NO(g) + 6H_2O(g) In a certain experiment, 1.85 g of NH_3 reacts with 3.39 g of O_2. Which is the limiting reactant? NH_3 O_2 How many grams of NO and of H_2O form? Enter your answers numerically separated by a comma. mNO, mH_2O = How many grams of the excess reactant remain after the limiting reactant is completely consumed? Express your answer in grams to three significant figures. M = g

Explanation / Answer

mol of NH3 = mass/MW = 1.85/17 = 0.1088

mol of O2 = masS/MW = 3.39/32 = 0.1059375

ratio is 4:5 so clearly NH3 is limiting

0.1088 mol of NH3 --> 0.1088 mol of NO 6/4*0.1088 = 0.1632 mol of H2O

mass of NO = 0.1088 *30 = 3.264 g

mass of H2O= 0.1632 *18= 2.9376 g

c.

excess left:

0.1059375 - 4/5*0.1088 = 0.0188975 mol of O2 left

mass = mol*MW = 0.0188975*32 = 0.60472 g of O2

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