# How do you balance RbNO_3 + BeF_2 -> Be(NO_3)_2 + RbF?

Dec 7, 2015

"RbNO"_text(3(aq]) + "BeF"_text(2(aq]) -> color(red)("N. R.")

#### Explanation:

The most important thing to notice here is that no reaction takes place when these two solutions are mixed.

The reason for why that happens lies with the fact that both products are soluble in aqueous solution, which implies that they will exist as dissociated ions.

Rubidium nitrate, ${\text{RbNO}}_{3}$, and beryllium fluoride, ${\text{BeF}}_{2}$, are soluble compounds that dissociate completely in aqueous solution to produce

${\text{RbNO"_text(3(aq]) -> "Rb"_text((aq])^(+) + "NO}}_{\textrm{3 \left(a q\right]}}^{-}$

and

${\text{BeF"_text(2(aq]) -> "Be"_text((aq])^(2+) + 2"F}}_{\textrm{\left(a q\right]}}^{-}$

In order for a double replacement reaction to take place, you need one of the products to be insoluble, i.e. precipitate out of solution. In this case, all the four compounds will exist as ions in solution, which means that the correct answer to this question is

"RbNO"_text(3(aq]) + "BeF"_text(2(aq]) -> color(red)("N. R.")

Here $\text{N"."R} .$ means "no reaction".

Now, assuming that these two compounds would actually react to form an insoluble compound, you can balance the equation by looking at ions, rather than at individual atoms.

Notice that you have a $\left(2 +\right)$ charge on the beryllium cation, but only a $\left(1 -\right)$ charge on the nitrate anion, ${\text{NO}}_{3}^{-}$. This means that you will need two nitrate anions to balance the positive charge of the beryllium cation.

To get two nitrate anions for every one beryllium cation, you need to multiply rubidium nitrate by $2$.

This will of course get you two rubidium cations as well. Notice that you have two fluoride anions, ${\text{F}}^{-}$, coming from beryllium fluoride, so you need to multiply rubidium fluoride by $2$ to make sure that both of the ions are balanced.

This would get you

$2 {\text{RbNO"_text(3(aq]) + "BeF"_text(2(aq]) -> "Be"("NO"_3)_text(2(aq]) + 2"RbF}}_{\textrm{\left(a q\right]}}$

Remember, this reaction does not take place, since all compounds are **soluble in aqueous solution.

$\textcolor{red}{\cancel{\textcolor{b l a c k}{2 {\text{RbNO"_text(3(aq]) + "BeF"_text(2(aq]) -> "Be"("NO"_3)_text(2(aq]) + 2"RbF}}_{\textrm{\left(a q\right]}}}}}$