Question

- Sodium peroxide reacts vigorously with water to produce sodium hydroxide and oxygen. Two questions: (a) What mass of O2 is produced when 60.0g of Na2O2 react? Check to make sure the equation is balanced.
2 Na2O2 + 2 H20 +4 NaOH + O2
Molar mass of known
Molar mass of the unknown
Possible Mole ratios
2 mol Na2O2 / 1 mol O2
1 mol O2 / 2 mol Na2O2
Molar mass
Na2O2 = (2*22.99)+(2*16.00)=77.98 g/mol
O2 = (2*16.00) = 32.00 g/mol

Answer 1

To determine the mass of O2 produced when 60.0 g of Na2O2 reacts with water, let's follow these steps:

1. **Check the balanced equation**:
The reaction provided is:
\[ 2 \text{Na}_2\text{O}_2 + 2 \text{H}_2\text{O} \rightarrow 4 \text{NaOH} + \text{O}_2 \]
This equation is balanced.

2. **Calculate the number of moles of Na2O2**:
Using the molar mass of Na2O2, which is 77.98 g/mol:
\[
\text{Moles of Na}_2\text{O}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{60.0 \, \text{g}}{77.98 \, \text{g/mol}} \approx 0.771 \text{ mol}
\]

3. **Use the mole ratio to find moles of O2 produced**:
From the balanced equation, 2 moles of Na2O2 produce 1 mole of O2. The mole ratio is:
\[
\frac{1 \text{ mole O}_2}{2 \text{ moles Na}_2\text{O}_2}
\]
Therefore, the moles of O2 produced from the moles of Na2O2 is:
\[
\text{Moles of O}_2 = 0.771 \, \text{mol Na}_2\text{O}_2 \times \frac{1 \, \text{mol O}_2}{2 \, \text{mol Na}_2\text{O}_2} = 0.3855 \, \text{mol O}_2
\]

4. **Calculate the mass of O2 produced**:
We can now find the mass of O2 produced using its molar mass (32.00 g/mol):
\[
\text{Mass of O}_2 = \text{moles} \times \text{molar mass} = 0.3855 \, \text{mol} \times 32.00 \, \text{g/mol} \approx 12.32 \, \text{g}
\]

So, the mass of O2 produced when 60.0 g of Na2O2 reacts is approximately **12.32 grams**.