Question

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

`"2Na"_2"O"_2 + "H"_2"O"` `rarr` `"4NaOH + O"_2"`

Answer 1

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

Explanation:

`"2Na"_2"O"_2"+2H"_2"O"` `rarr` `"4NaOH"+"O"_2"`

Determine the mole ratios between `"Na"_2"O"_2` and `"O"_2"` from the balanced equation.

`("2 mol Na"_2"O"_2)/("1 mol O"_2")` and `("1 mol O"_2)/("2 mol Na"_2"O"_2")`

Determine the molar masses of `"Na"_2"O"_2` and `"O"_2`.

`"Na"_2"O"_2:` `(2xx22.990)+(2xx15.999)="77.978 g/mol"`

`"O"_2:` `(2xx15.999)="31.998 g/mol"`

Determine the number of moles of `"Na"_2"O"_2` in `"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 `"O"_2"` that can be produced by `"0.014748 mol Na"_2"O"_2` by multiplying the moles of `"Na"_2"O"_2` times the mole ratio with `"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 `"0.0073740 mol O"_2"` by multiplying the moles `"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"`

Answer 2

First balance the equation to find that so the moles of `O_2` released is equal to one-half the moles of `Na_2O_2` added:
`2 Na_2O_2 + 2 H_2O rarr 4 NaOH + O_2`

Explanation:

The molar mass of `Na_2O_2` is 77.98 g/mol, so the moles of `Na_2O_2` is
`(1.15g)/(77.98 g/(mol))=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
`1/2 times 0.01456 mol times 32.00 g/(mol)=0.233`g