Question

downloaded from stanmorephysics.com
grade 11 - ncs
question 2 (start on a new page)
ozone (o₃) reacts with nitrogen monoxide (no) to produce no₂ gas according to the following balanced equation:
o₃(g) + no(g) → o₂(g) + no₂(g)
in the reaction, 1,5 g of o₃ reacts with 1,25 g of no.
2.1 define the term limiting reactant.
2.2 determine a substance that is a limiting reactant in the reaction.
2.3 calculate the:
2.3.1 mass of no₂(g) produced.
2.3.2 mass of excess reactant when the reaction is complete.

Explanation:

2.1

A limiting reactant is a reactant in a chemical reaction that is completely consumed first, thereby limiting the amount of product that can be formed because the reaction stops once this reactant is used up.

Step1: Calculate moles of \( O_3 \)

Molar mass of \( O_3 = 3\times16 = 48 \, g/mol \). Moles of \( O_3 = \frac{mass}{molar\ mass} = \frac{1.5\ g}{48\ g/mol} = 0.03125\ mol \).

Step2: Calculate moles of \( NO \)

Molar mass of \( NO = 14 + 16 = 30 \, g/mol \). Moles of \( NO = \frac{1.25\ g}{30\ g/mol} \approx 0.04167\ mol \).

Step3: Compare mole ratio

From the balanced equation \( O_3 + NO
ightarrow O_2 + NO_2 \), the mole ratio of \( O_3:NO \) is \( 1:1 \). Moles of \( O_3 (0.03125\ mol) \) are less than moles of \( NO (0.04167\ mol) \), so \( O_3 \) is the limiting reactant.

Step1: Moles of \( NO_2 \) from limiting reactant

Since \( O_3 \) is limiting and mole ratio \( O_3:NO_2 = 1:1 \), moles of \( NO_2 = \) moles of \( O_3 = 0.03125\ mol \).

Step2: Molar mass of \( NO_2 \)

Molar mass of \( NO_2 = 14 + 2\times16 = 46 \, g/mol \).

Step3: Calculate mass of \( NO_2 \)

Mass = moles × molar mass = \( 0.03125\ mol \times 46\ g/mol = 1.4375\ g \).

Answer:

A limiting reactant is a reactant in a chemical reaction that is completely consumed, thereby limiting the amount of product formed as the reaction ceases when this reactant is exhausted.

2.2