# If you start with 5.5 grams of sodium fluoride, how many grams of magnesium fluoride will be produced in the reaction Mg + 2NaF -> MgF_2 + 2Na?

Apr 28, 2016

#### Answer:

There will be 4.1 g of ${\text{MgF}}_{2}$ produced.

#### Explanation:

Mass of sodium fluoride = 5.5 g

$\text{moles" = "mass"/"molar mass}$

$\text{moles" = "5.5 g"/"(22.99 + 18.998) g/mole}$

$\text{moles" = 5.5/41.998 "moles}$

$\text{moles" = "0.13096 moles of NaF}$

For every 2 moles of $\text{NaF}$, there is 1 mole of ${\text{MgF}}_{2}$.

Using this knowledge, we can determine the amount of moles of ${\text{MgF}}_{2}$.

${\text{0.13096 moles of NaF"/2 = "moles of MgF}}_{2}$

$\text{moles of MgF"_2 = "0.06548 moles}$

We are now going to convert this into grams.

$\text{moles" = "mass"/"molar mass}$

$\text{0.06548 moles" = "mass"/"(24.305+2(18.998)) g/mole}$

$0.06548 = \text{mass"/"62.301 g}$

$\text{mass of MgF"_2 = "4.1 grams}$

Therefore, there will be 4.1 grams of ${\text{MgF}}_{2}$ produced.