What is the mass of heavy water, D2O (l), produced when 7.60 g of O2 (g) reacts with excess D2 (g) ?

19.026 g

Explanation:

The reaction between D₂ and O₂ is given by the equation;

2D₂ (g) + O₂ (g) → 2H₂O (l)

We are given;

Mass of oxygen (O₂) that reacted as 7.60 g

We are required to calculate the mass of D₂O produced;

Step 1: Calculate the moles of O₂ used (limiting reactant)

To calculate the number of moles we divide mass by the molar mass.

Moles = mass : Molar mass

Molar mass of O₂ is 16.0 g/mol

Therefore;

Moles = 7.60 g : 16.0 g/mol

= 0.475 moles

Step 2: Calculate the moles of D₂O produced

From the equation, 1 mole of oxygen reacts to produce 2 moles of D₂O

Thus, the mole ratio of O₂ : D₂O is 1 : 2

Therefore, moles of D₂O = Moles of O₂ * 2

= 0.475 moles * 2

= 0.95 moles

Step 3: Calculate the mass of heavy water produced

Mass = Number of moles * Molar mass

Molar mass of heavy water = 20.0276 g/mol

Therefore;

Mass of heavy water = 0.95 moles * 20.0276 g/mol

= 19.026 g

Hence, the mass of heavy water produced is 19.026 g