# 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?

Dec 13, 2015

The amount of $\text{O"_2}$ consumed is 6.83 g.

#### Explanation:

Balanced Equation

"2Mg(s)" + "O"_2("g")$\rightarrow$$\text{2MgO(s)}$

You need the molar masses of $\text{MgO}$ and $\text{O"_2}$.

Molar Masses
$\text{MgO} :$$\text{40.304400 g/mol}$
http://www.ncbi.nlm.nih.gov/pccompound?term=%22Magnesium+oxide+(MgO)%22

${\text{O}}_{2} :$$\text{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.

$17.2 \cancel{\text{g MgO"xx(1"mol MgO")/(40.304400cancel"g MgO")="0.42675 mol MgO}}$

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 $\text{O"_2}$ and $\text{MgO}$ from the balanced equation. This will give the moles of $\text{O"_2}$ that were consumed.

$0.42675 \cancel{\text{mol MgO"xx(1"mol O"_2)/(2cancel"mol MgO")="0.21338 mol O"_2}}$

Multiply the moles of $\text{O"_2}$ times the molar mass of $\text{O"_2}$. This will give the mass of $\text{O"_2}$ consumed.

$0.21338 \cancel{\text{mol O"_2xx(31.9988"g O"_2)/(1cancel"mol O"_2)="6.83 g O"_2}}$

You can combine all of the steps as shown below.

$17.2 \cancel{\text{g MgO"xx(1cancel"mol MgO")/(40.304400cancel"g MgO")xx(1cancel"mol O"_2)/(2cancel"mol MgO")xx(31.9988"g O"_2)/(1cancel"mol O"_2)="6.83 g O"_2}}$