Question

17.2 grams `MgO` are produced according to the following equation: `2Mg + O_2 -> 2MgO`. How many grams of oxygen `O_2` must have been consumed?

Answer 1

The amount of `"O"_2"` consumed is 6.83 g.

Explanation:

Balanced Equation

`"2Mg(s)" + "O"_2("g")` `rarr` `"2MgO(s)"`

You need the molar masses of `"MgO"` and `"O"_2"`.

Molar Masses
`"MgO":` `"40.304400 g/mol"`
http://www.ncbi.nlm.nih.gov/pccompound?term=%22Magnesium+oxide+(MgO)%22

`"O"_2:` `"31.9988 g/mol"`
http://pubchem.ncbi.nlm.nih.gov/compound/977#section=Top

Determine the moles of MgO by dividing the mass of MgO produced by its molar mass.

`17.2cancel"g MgO"xx(1"mol MgO")/(40.304400cancel"g MgO")="0.42675 mol MgO"`

I am keeping some guard digits to reduce rounding errors. I will round to three significant figures at the end.

Multiply the moles of MgO times the mole ratio between `"O"_2"` and `"MgO"` from the balanced equation. This will give the moles of `"O"_2"` that were consumed.

`0.42675cancel"mol MgO"xx(1"mol O"_2)/(2cancel"mol MgO")="0.21338 mol O"_2"`

Multiply the moles of `"O"_2"` times the molar mass of `"O"_2"`. This will give the mass of `"O"_2"` consumed.

`0.21338cancel"mol O"_2xx(31.9988"g O"_2)/(1cancel"mol O"_2)="6.83 g O"_2"`

You can combine all of the steps as shown below.

`17.2cancel"g MgO"xx(1cancel"mol MgO")/(40.304400cancel"g MgO")xx(1cancel"mol O"_2)/(2cancel"mol MgO")xx(31.9988"g O"_2)/(1cancel"mol O"_2)="6.83 g O"_2"`