The question is incomplete, the complete question is;
One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO How many grams of NO and of H20 form? Enter your answers numerically separated by a comma. 4NH3(g) +502(g)------->4NO(g)+6H2O(g)
In a certain experiment, 1.10 g of NH3 reacts with 2.02 g of O2. How many grams of the excess reactant remain after the limiting reactant is completely consumed? Express your answer in grams to three significant figures.
Answer:
Mass of excess ammonia 0.034 g of ammonia
Mass of water formed= 1.37g
Mass of NO formed = 1.50g
Explanation:
The limiting reactant is the reactant that yields the least number of moles of product.
For NH3, molar mass of ammonia = 17g mol-1
Number of moles of ammonia reacted= 1.10g/17 gmol-1 = 0.065 moles of ammonia
According to the reaction equation;
4 moles of ammonia yields 4 moles of NO
Hence 0.065 moles of ammonia will yield 0.065 ×4/4 = 0.065 moles of NO
For oxygen, molar mass of oxygen gas = 32gmol-1
Number of moles of oxygen gas= 2.02g/32gmol-1 = 0.063 moles of oxygen
From the reaction equation;
5 moles of oxygen gas yields 4 moles of NO
0.063 moles of oxygen will yield 0.063 ×4 /5 = 0.050 moles of NO
Hence oxygen is the limiting reactant and ammonia is the excess reactant.
Amount of excess ammonia = Amount of ammonia - amount of oxygen
Amount of excess ammonia= 0.065-0.063= 2×10^-3 moles
Mass of excess ammonia = 2×10^-3 moles × 17 gmol-1 = 0.034 g of ammonia
Mass of NO formed is obtained from the limiting reactant. Since molar mass of is 30gmol-1. Then mass of NO formed = 0.050 moles of NO × 30gmol-1 = 1.50 g of NO
For water;
5 moles of oxygen yields 6 moles of water
Hence 0.063 moles of oxygen yields 0.063 × 6/5 = 0.076 moles of water
Molar mass of water = 18gmol-1
Hence mass of water = 0.076 moles × 18gmol-1 = 1.37g of water
