Take the reaction : NH3 + O2 = NO + H2O
in a experiment, 5 moles of NH3 are alloved to react with 5 moles of O2
a) which reactant is the limiting reagent ?
b) How many grams of NO are formed ?
c) How much of the excess reactant remains after the reaction ?
To determine the limiting reagent in a chemical reaction, you need to compare the moles of each reactant to the stoichiometric ratios. The reactant that produces fewer moles of the desired product is the limiting reagent.
a) Firstly, let's determine the stoichiometric ratio of moles of NH3 and O2 in the balanced equation:
NH3 + O2 = NO + H2O
From the balanced equation, we can see that 1 mole of NH3 reacts with 1 mole of O2 to produce 1 mole of NO.
Given that 5 moles of NH3 and 5 moles of O2 are available, we compare the number of moles available with the stoichiometric ratios. Since both reactants have the same number of moles, the limiting reagent cannot be determined in this case. The reaction will go to completion with both reactants being fully consumed.
b) To calculate the number of grams of NO formed, we need to determine the molar mass of NO and convert moles to grams.
The molar mass of NO is calculated as follows:
N = 14.01 g/mol
O = 16.00 g/mol
Molar mass of NO = N + O = 14.01 g/mol + 16.00 g/mol = 30.01 g/mol
Since we have established that both NH3 and O2 are fully consumed, we can calculate the number of moles of NO produced by taking the minimum number of moles from the two reactants, which is 5 moles.
Mass of NO = number of moles × molar mass
Mass of NO = 5 moles × 30.01 g/mol = 150.05 g
Therefore, 150.05 grams of NO are formed.
c) Since we established that both NH3 and O2 are fully consumed, there will be no excess reactant remaining after the reaction.