Question

How many grams of Ca does 32 grams of O2 (oxygen)have to react to give 72 grams of CaO? Answer is given 40 grams but if 32 grams of oxygen has to react then it means that there are two oxygen atoms and then the formula will be CaO2

Answer 1

So someone made a mistake...

Explanation:

The stoichiometric equation is...

`Ca(s) + 1/2O_2(g) rarr CaO(g)`

And so ONE equiv of metal reacts with an 1/2 EQUIV of dioxygen gas...

You got `72*g` calcium oxide, a molar quantity of `(72*g)/(56.08*g*mol^-1)=1.28*mol`...

And SO WITH RESPECT TO OXYGEN there is a mass of ...

`1.28*molxx16.00*g*mol^-1=20.5*g`....

Answer 2

The given answer is incorrect.

Calcium is the limiting reagant, and `"51 g Ca"` will react with `"32 g O"_2"` to produce `"72 g CaO"`.

Oxygen gas is in excess. Only `"21 g O"_2` are needed to produce `"72 g CaO"`. (These are approximations due to the use of just two significant figures.)

Even if `"100 g O"_2"` were available and it reacted with `"51 g Ca"`, there would still be only `"72 g CaO"` produced.

Explanation:

We need to determine how much `"Ca"` will produce `"72 g CaO"`.

We need the molar masses of each reactant and the product.

`"O"_2":` `"31.998 g/mol"`

`"Ca":` `"40.078 g/mol"`

`"CaO":` `"56.077 g/mol"`

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The processes:

Mass of Ca required to produce 72 g CaO.

`"mass CaO"` `rarr` `"mol CaO"` `rarr` `"mol Ca"` `rarr` `"mass Ca"`.

Mass of CaO from 32 g `"O"_2`

`"mass O"_2"` `rarr` `"mol O"_2"` `rarr` `"mol CaO"` `rarr` `"mass CaO"`

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Determine how much calcium is needed to produce `"72 g CaO"`

Moles CaO from 72 g CaO

Multiply given mass `"CaO"` by the reciprocal of its molar mass, `"mol"/"g"`.

`72color(red)cancel(color(black)("g CaO"))xx(1"mol CaO")/(56.077color(red)cancel(color(black)("g CaO")))="1.2839 mol CaO"`

I am keeping some extra digits to reduce rounding errors. The final answer will be rounded to two significant figures.

Moles Ca from moles CaO

Multiply mol CaO by the mole ratio between Ca and CaO from the balanced equation, with Ca in the numerator.

`1.2839color(red)cancel(color(black)("mol CaO"))xx(2"mol Ca")/(2color(red)cancel(color(black)("mol CaO")))="1.2839 mol Ca"`

Mass Ca from moles Ca

Multiply moles Ca by its molar mass.

`1.2839color(red)cancel(color(black)("mol Ca"))xx(40.078"g Ca")/(1color(red)cancel(color(black)("mol Ca")))="51 g Ca"` (rounded to two significant figures)

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Mass of `"CaO"` from `"O"_2"`

Moles oxygen gas from 32 g `"O"_2`

Multiply the given mass by the reciprocal of its molar mass, `"mol"/"g"`.

`32color(red)cancel(color(black)("g O"_2))xx(1"mol O"_2)/(31.998color(red)cancel(color(black)("g O"_2)))="1.0 mol O"_2`

Moles CaO from moles `"O"_2"`

Multiply mol `"O"_2"` by the mole ratio between `"CaO"` and `"O"_2"` from the balanced equation, with `"CaO"` in the numerator.

`1.0color(red)cancel(color(black)("mol O"_2))xx(2"mol CaO")/(1color(red)cancel(color(black)("mol O"_2)))="2.0 mol CaO"`

Mass CaO from moles CaO

Multiply moles CaO by its molar mass.

`2.0color(red)cancel(color(black)("mol CaO"))xx(56.077"g CaO")/(1color(red)cancel(color(black)("mol CaO")))="110 g CaO"` (rounded to two significant figures)

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Calcium is the limiting reagant, and `"51 g Ca"` will react with `"32 g O"_2"` to produce `"72 g CaO"`.

Oxygen gas is in excess.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Amount of excess `"O"_2`.

Because of the law of conservation of mass, we know that the mass of the reactants must equal the mass of the products. So in order to determine the actual mass of `"O"_2` that can react with `"51 g Ca"` to produce `"72 g CaO"`, we can subtract the mass of `"CaO"` from `"72 g CaO"`

`"72 g CaO"` `-` `"51 g Ca"="21 g O"_2`

The amount of oxygen in excess is:

`"32 g O"_2"` `-` `"21 g O"_2"` `=` `"11 g O"_2`