In a reaction between water and hydrogen to form hydrogen peroxide , determine the mass of unused oxygen if 1 1/2 moles of water and 1 mole of oxygen are mixed up

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Solution

1. The reaction is as follows :
H₂O (aq) + O₂ (l) $\to$ H₂O₂(aq)
2. 1 mole of water combines with 1 mole of O₂.
3. So,half moles of oxygen remains unused.
4. Mass of 1 mole of O₂ = 32g
Mass of 0.5 mole of O₂ = 16g.
5. So,Mass of unused O₂ = 16g

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In a reaction between water and hydrogen to form hydrogen peroxide , determine the mass of unused oxygen if 1 1/2 moles of water and 1 mole of oxygen are mixed up

1. The reaction is as follows : H2O (aq) + O2 (l) → H2O2(aq) 2. 1 mole of water combines with 1 mole of O2. 3. So,half moles of oxygen remains unused. 4. Mass of 1 mole of O2 = 32g Mass of 0.5 mole of O2 = 16g. 5. So,Mass of unused O2 = 16g

1. The reaction is as follows :
H₂O (aq) + O₂ (l) $\to$ H₂O₂(aq)
2. 1 mole of water combines with 1 mole of O₂.
3. So,half moles of oxygen remains unused.
4. Mass of 1 mole of O₂ = 32g
Mass of 0.5 mole of O₂ = 16g.
5. So,Mass of unused O₂ = 16g