Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a
dark-brown gas:
2NO(g) + O2(g) -> 2NO2(g)
In one experiment 0.886 moles of NO is mixed with 0.503 moles of O2.
Calculate which of the two reactants is the limiting reagent.
Please show set up of problem and show work.
To determine which reactant is the limiting reagent, we need to compare the stoichiometric ratios of the reactants to the reaction.
Given:
moles of NO = 0.886 mol
moles of O2 = 0.503 mol
First, we'll calculate the moles of NO2 that can be formed from each reactant based on the stoichiometric ratios of the balanced equation.
From the balanced equation:
2 moles of NO reacts with 1 mole of O2 to produce 2 moles of NO2.
Moles of NO2 from NO = (0.886 mol of NO) x (2 mol of NO2 / 2 mol of NO) = 0.886 mol of NO2
Moles of NO2 from O2 = (0.503 mol of O2) x (2 mol of NO2 / 1 mol of O2) = 1.006 mol of NO2
Now, we compare the moles of NO2 obtained from each reactant.
Moles of NO2 from NO = 0.886 mol
Moles of NO2 from O2 = 1.006 mol
Since the moles of NO2 obtained from NO (0.886 mol) is less than the moles of NO2 obtained from O2 (1.006 mol), the limiting reagent is NO.
To determine which of the two reactants is the limiting reagent, you need to compare the number of moles of each reactant to the stoichiometric coefficients of the balanced chemical equation.
Given:
Moles of NO = 0.886 mol
Moles of O2 = 0.503 mol
Step 1: Determine the mole ratio between the two reactants.
Looking at the balanced equation: 2NO(g) + O2(g) -> 2NO2(g), the stoichiometric coefficient for NO is 2, and for O2 it is 1.
Step 2: Calculate the moles of NO that would react with the given amount of O2.
To do this, multiply the moles of O2 by the mole ratio of NO to O2.
Moles of NO reacted = 0.503 mol O2 × (2 mol NO / 1 mol O2) = 1.006 mol NO
Step 3: Compare the calculated moles of NO with the actual moles of NO.
If the calculated moles of NO are less than the actual moles of NO, then O2 is the limiting reagent. However, if the calculated moles of NO are greater than or equal to the actual moles of NO, then NO is the limiting reagent.
In this case, the calculated moles of NO (1.006 mol) are greater than the actual moles of NO (0.886 mol). Therefore, NO is the limiting reagent.