# What mass of oxygen gas can be be produced when 1.15 g sodium peroxide reacts with excess water?

## $\text{2Na"_2"O"_2 + "H"_2"O}$$\rightarrow$$\text{4NaOH + O"_2}$

Jul 3, 2015

The mass of oxygen that be produced from 1.15 g sodium peroxide is 0.236 g.

#### Explanation:

$\text{2Na"_2"O"_2"+2H"_2"O}$$\rightarrow$$\text{4NaOH"+"O"_2}$

Determine the mole ratios between ${\text{Na"_2"O}}_{2}$ and $\text{O"_2}$ from the balanced equation.

$\left(\text{2 mol Na"_2"O"_2)/("1 mol O"_2}\right)$ and $\left(\text{1 mol O"_2)/("2 mol Na"_2"O"_2}\right)$

Determine the molar masses of ${\text{Na"_2"O}}_{2}$ and ${\text{O}}_{2}$.

${\text{Na"_2"O}}_{2} :$$\left(2 \times 22.990\right) + \left(2 \times 15.999\right) = \text{77.978 g/mol}$

${\text{O}}_{2} :$$\left(2 \times 15.999\right) = \text{31.998 g/mol}$

Determine the number of moles of ${\text{Na"_2"O}}_{2}$ in $\text{1.15 g}$ using its molar mass.

1.15color(red)cancel(color(black)("g Na"_2"O"_2))xx(1"mol Na"_2"O"_2)/(77.978color(red)cancel(color(black)("g Na"_2"O"_2)))="0.014748 mol Na"_2"O"_2"

Determine the number of moles of $\text{O"_2}$ that can be produced by ${\text{0.014748 mol Na"_2"O}}_{2}$ by multiplying the moles of ${\text{Na"_2"O}}_{2}$ times the mole ratio with ${\text{O}}_{2}$ in the numerator.

0.014748 color(red)cancel(color(black)("mol Na"_2"O"_2))xx("1 mol O"_2)/(2 color(red)cancel(color(black)("mol Na"_2"O"_2)))="0.0073740 mol O"_2"

Determine the mass in grams in $\text{0.0073740 mol O"_2}$ by multiplying the moles $\text{O"_2}$ times its molar mass.

0.0073740 color(red)cancel(color(black)("mol O"_2))xx(31.998"g O"_2)/(1color(red)cancel(color(black)("mol O"_2)))="0.236 g O"_2"

Jul 4, 2015

First balance the equation to find that so the moles of ${O}_{2}$ released is equal to one-half the moles of $N {a}_{2} {O}_{2}$ added:
$2 N {a}_{2} {O}_{2} + 2 {H}_{2} O \rightarrow 4 N a O H + {O}_{2}$

#### Explanation:

The molar mass of $N {a}_{2} {O}_{2}$ is 77.98 g/mol, so the moles of $N {a}_{2} {O}_{2}$ is
$\frac{1.15 g}{77.98 \frac{g}{m o l}} = 0.01456$ mol
(carries 1 extra sig fig for intermediate calculation)

The molar mass of ${O}_{2}$ is 32.00 g/mol, so the mass of ${O}_{2}$ released is
$\frac{1}{2} \times 0.01456 m o l \times 32.00 \frac{g}{m o l} = 0.233$g