Question

Using the image seen here, select the best answer for the sample reaction. Some of the information is provided to get you started with things you might need for the calculation.

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 H2O------> 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



(1 point)
Responses

109.98 g
109.98 g

32.00 g
32.00 g

117.50 g
117.50 g

12.31 g

Answer 1

To find the mass of O2 produced when 60.0 g of Na2O2 reacts, we can follow these steps:

1. **Determine the moles of Na2O2** using its molar mass:
\[
\text{Molar mass of Na2O2} = 77.98 \text{ g/mol}
\]

\[
\text{Moles of Na2O2} = \frac{\text{mass}}{\text{molar mass}} = \frac{60.0 \text{ g}}{77.98 \text{ g/mol}} \approx 0.769 \text{ moles}
\]

2. **Use the mole ratio from the balanced equation** to find the moles of O2 produced:
From the balanced equation:
\[
2 \text{ moles of Na2O2} \rightarrow 1 \text{ mole of O2}
\]
Therefore:
\[
\text{Moles of O2} = \frac{0.769 \text{ moles Na2O2}}{2} = 0.3845 \text{ moles of O2}
\]

3. **Calculate the mass of O2 produced**:
\[
\text{Molar mass of O2} = 32.00 \text{ g/mol}
\]
\[
\text{Mass of O2} = \text{moles} \times \text{molar mass} = 0.3845 \text{ moles} \times 32.00 \text{ g/mol} \approx 12.31 \text{ g}
\]

Thus, the correct answer is **12.31 g**.