Question

Nitric oxide (NO) reacts with oxygen gas to produce nitrogen dioxide. A gaseous mixture contains 0.66 g of nitric oxide and 0.58 g of oxygen gas. After the reaction is complete, what mass of nitrogen dioxide is formed? Which reactant is in excess? How do you know? Suppose you actually recovered 0.91 g of nitrogen dioxide. What is your percent yield?

Answer 1

The mass of nitrogen dioxide formed after nitric oxide reacts with oxygen gas is 1.01 g, and oxygen gas is the reactant in excess. The percent yield of the reaction, given that 0.91 g of nitrogen dioxide is actually recovered, is 90.1%.

Step-by-step explanation:

When nitric oxide (NO) reacts with oxygen gas (O₂) to produce nitrogen dioxide (NO₂), we can use stoichiometry to determine the mass of the product formed and the reactant in excess. First, we need to balance the chemical equation:

2 NO(g) + O₂(g) → 2 NO₂(g)

Next, we convert the masses of the reactants into moles using their molar masses (NO = 30.01 g/mol, O₂ = 32.00 g/mol):

• 0.66 g NO ÷ 30.01 g/mol = 0.022 moles NO

• 0.58 g O₂ ÷ 32.00 g/mol = 0.018 moles O₂

From the balanced equation, 1 mole of O₂ reacts with 2 moles of NO, hence the limiting reactant is NO since it will run out first. The mass of NO₂ produced can be calculated:

• 0.022 moles NO × (46.01 g/mol NO₂) × (1 mole NO₂ / 1 mole NO) = 1.01 g NO₂

O₂ is in excess because we used less than the stoichiometric amount required to react with all the NO. To calculate the percent yield, we compare the actual yield to the theoretical yield:

• Percent Yield = (Actual Yield / Theoretical Yield) × 100

• Percent Yield = (0.91 g / 1.01 g) × 100 = 90.1%

Answer 2

NO is the limiting reagent.

In this reaction 0.886 mole of NO2 is produced

Step-by-step explanation:

The chemical equation for this reaction is

2NO(g) + O2(g) → 2NO2(g)

In this limiting reagent reaction, 2 moles of NO reacts with one mole of O2 to produce 2 mole of 2NO2

0.886 mole of NO * (2 mole of NO2/2 mole of NO) = 0.886 mole of NO2

0.503 mole of O2 * (1 mole of NO2/1 mole of O2) = 1.01 mole of NO2

Hence, NO is the limiting reagent.

In this reaction 0.886 mole of NO2 is produced