# Question 176ac

Mar 10, 2016

$\text{97.2 g Mg}$

#### Explanation:

The first thing to do here is make sure that you write a balanced chemical equation for this synthesis reaction.

Magnesium, $\text{Mg}$ is located in group 2 of the periodic table, which should automatically tell you that it must lose two electrons to completes its octet.

Oxygen, on the other hand, is located in group 16 of the periodic table, so it needs two electrons to complete its octet.

When the two elements react, magnesium will donate its two valence electrons to oxygen to form ${\text{Mg}}^{2 +}$, and oxygen will pick up those two valence electrons to form ${\text{O}}^{2 -}$.

The chemical formula for magnesium oxide will thus be $\text{MgO}$.

The balanced chemical reaction looks like this

$\textcolor{red}{2} {\text{Mg"_text((s]) + "O"_text(2(g]) -> 2"MgO}}_{\textrm{\left(s\right]}}$

Notice that the reaction consumes $\textcolor{red}{2}$ moles of magnesium for every $1$ mole of oxygen gas - this means that you have a $\textcolor{red}{2} : 1$ mole ratio between the two reactants.

Now, the problem tells you that you have $2$ moles of oxygen gas available for the reaction.

Well, if you need two moles of magnesium for every one mole of oxygen gas, it follows that this reaction will need

2.00color(red)(cancel(color(black)("moles O"_2))) * overbrace((color(red)(2)color(white)(a)"moles Mg")/(1color(red)(cancel(color(black)("mole O"_2)))))^(color(purple)("2:1 mole ratio")) = "4.00 moles Mg"#

All you need to do now is use the mass of one mole of magnesium to determine how many grams would contain 4 moles of magnesium.

This value is given by the element's molar mass, which tells you the mass of one mole of magnesium atoms.

Magnesium has a molar mass of ${\text{24.305 g mol}}^{- 1}$, so one mole of magnesium has a mass of $\text{24.305 g}$. This means that you have

$4.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Mg"))) * overbrace("24.305 g"/(1color(red)(cancel(color(black)("mole Mg")))))^(color(brown)("molar mass of Mg")) = color(green)(|bar(ul(color(white)(a/a)"97.2 g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the number of moles of oxygen gas.