# What volume of oxygen can be obtained by the addition of excess water to 1.0g of sodium peroxide?

Oct 23, 2015

$\text{0.15 L } \to$ at STP

#### Explanation:

First thing first, you can't calculate the volume of oxygen produced by the reaction if you don't know the conditions for pressure and temperature.

SInce no mention of this was made, I'll assume that the reaction takes place at STP - Standard Temperature and Pressure, which implies a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

Now focus on writing the balanced chemical equation for this reaction.

Sodium peroxide, ${\text{Na"_2"O}}_{2}$, will react with water to form sodium hydroxide, $\text{NaOH}$, and hydrogen peroxide, ${\text{H"_2"O}}_{2}$.

${\text{Na"_2"O"_text(2(s]) + 2"H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + "H"_2"O}}_{\textrm{2 \left(a q\right]}}$

The oxygen gas will actually come from the decomposition of the hydrogen peroxide.

$2 {\text{H"_2"O"_text(2(aq]) -> 2"H"_2"O"_text((l]) + "O}}_{\textrm{2 \left(g\right]}} \uparrow$

This equation can be rewritten as

${\text{H"_2"O"_text(2(aq]) -> "H"_2"O"_text((l]) + 1/2"O}}_{\textrm{2 \left(g\right]}} \uparrow$

Combine these two equations to get the overall reaction

${\text{Na"_2"O"_text(2(s]) + color(red)(cancel(color(black)(2)))"H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + color(red)(cancel(color(black)("H"_2"O"_text((l])))) + 1/2"O}}_{\textrm{2 \left(g\right]}} \uparrow$

${\text{Na"_2"O"_text(2(s]) + "H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + 1/2"O}}_{\textrm{2 \left(g\right]}} \uparrow$

Notice that the reaction produces $\frac{1}{2}$ moles of oxygen gas for every mole of sodium peroxide.

To determine how many moles of sodium peroxide you have, use the compound's molar mass

1.0color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"O"_2)/(77.978color(red)(cancel(color(black)("g")))) = "0.0128 moles Na"_2"O"_2

The reaction will produce

0.0128color(red)(cancel(color(black)("moles Na"_2"O"_2))) * ("1/2 moles O"_2)/(1color(red)(cancel(color(black)("mole Na"_2"O"_2)))) = "0.00640 moles O"_2

At STP, one mole of any ideal gas occupies exactly $\text{22.7 L}$ - this is known as the molar volume of a gas at STP.

In your case, this many moles of oxygen gas would occupy

0.00640color(red)(cancel(color(black)("moles"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.15 L")