Question

Nitric oxide reacts with oxygen gas to form nitrogen dioxide, a dark-brown gas 2NO+O₂→2NO₂.In one experiment, 0.866 mole of NO is mixed with 0.503 mole of O_2. Calculate which of the two reactants is the limiting reagent. Calculate also the number of NO2 produced.

a) NO is limiting; 0.723 mol NO2 produced
b) O_2 is limiting; 0.723 mol NO2 produced
c) NO is limiting; 1.446 mol NO2 produced
d) O_2 is limiting; 1.446 mol NO2 produced

Answer 1

O₂ is the limiting reagent and 1.006 mol of NO₂ is produced

Step-by-step explanation:

To determine which reactant is the limiting reagent, we need to compare the mole ratios of the reactants to the stoichiometry of the balanced equation. In this case, we have 2 moles of NO reacting with 1 mole of O₂ to produce 2 moles of NO₂.

Using the given amounts, we can calculate the moles of NO and O₂ in the reaction mixture.

0.866 mole of NO / 2 = 0.433 mole of NO and 0.503 mole of O₂ / 1 = 0.503 mole of O₂.

Since the mole ratio of NO to O₂ is 2:1, and we have less O₂ than the ratio requires, O₂ is the limiting reagent.

To calculate the number of moles of NO₂ produced, we can multiply the moles of the limiting reagent (O₂) by the mole ratio of NO₂ to O₂ (2:1).

0.503 mole of O₂ * (2 mole of NO₂ / 1 mole of O₂) = 1.006 mole of NO₂.

Therefore, the answer is option d) O₂ is limiting; 1.006 mol NO₂ produced.