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4 years ago

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

Chemistry

1 answer:

Rina8888 [55]4 years ago

3 0

<h3>Answer:</h3>

19.026 g

<h3>Explanation:</h3>

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;

<h3>Step 1: Calculate the moles of O₂ used (limiting reactant)</h3>

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

<h3>Step 2: Calculate the moles of D₂O produced </h3>

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

<h3>Step 3: Calculate the mass of heavy water produced</h3>

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

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the energy of the third excited rotational state $\mathbf{E_3 = 16.041 \ meV}$

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