Question

5. Sodium peroxide reacts vigorously with water in the following unbalanced equation:

Na202 + H20+ NaOH + O2
What mass of O2 is produced when 50.0 g Na2O2 reacts with H2O?
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Answer 1

To find the mass of O₂ produced when Na₂O₂ reacts with H₂O calculate the moles of Na₂O₂, then use the balanced equation to find the moles of O₂, and finally convert moles of O₂to grams using the molar mass of O₂. In this case, 29.1 grams of O₂are produced when 50.0 grams of Na₂O₂ reacts with H₂O.

Step-by-step explanation:

To determine the mass of O₂ produced when 50.0 g of Na₂O₂ reacts with H₂O, we need to calculate the number of moles of Na₂O₂ and use the balanced equation to find the moles of O₂. Then, we can convert the moles of O₂ to grams using the molar mass of O₂.

Step 1: Calculate the moles of Na₂O₂

moles = mass / molar mass

moles = 50.0 g / (22.99 g/mol + 2 * 16.00 g/mol)

moles = 50.0 g / 54.99 g/mol = 0.909 mol

Step 2: Use the balanced equation to find the moles of O₂

From the balanced equation Na₂O₂ + H₂O → 2 NaOH + O₂, we can see that 1 mole of Na₂O₂ reacts to form 1 mole of O₂.

moles of O₂ = 0.909 mol

Step 3: Convert moles of O₂ to grams using the molar mass of O₂

mass = moles * molar mass

mass = 0.909 mol * 32.00 g/mol = 29.1 g

Therefore, 29.1 grams of O2 are produced when 50.0 g of Na₂O₂ reacts with H₂O.

Answer 2

30.7 g of O₂ are produced

Step-by-step explanation:

This is a redox reaction, where the peroxide is reduced to oxygen and oxidized to hydroxide. The equation is this one:

2Na₂O₂ + 2H₂O → 4NaOH + 3O₂

We assume that water is in excess, so the peroxide is the limiting reagent.

We convert the mass to moles: 50 g . 1mol / 78 g = 0.641 moles

2 moles of peroxide can produce 3 moles of oxygen (according to stoichiometry)

Then, 0.641 moles of peroxide will produce (0.641 . 3 ) /2 = 0.961 moles O₂

Finally, we convert the moles to mass: 0.961mol . 32 g / 1mol = 30.7 g of O₂