# You have 18.0 g of aluminum metal, and a sample of 1.3M nitric acid in water. What volume of solution would you need to add to the metal to completely consume the aluminum according to the reaction below? 6 HNO3(aq) + 2Al(s) → 2 Al(NO3)3(aq)+ 3 H2(g)

Mar 14, 2016

$\text{1.5 L}$

#### Explanation:

Your strategy here will be to use the mole ratio that exists between the two reactants to determine how many moles of nitric acid are needed in order for all the moles of aluminium to take part in the reaction.

Once you know this, use the molarity of the nitric acid solution to determine what volume would contain that many moles of acid.

So, the balanced chemical equation for this single replacement reaction looks like this

$\textcolor{red}{2} {\text{Al"_text((s]) + color(purple)(6)"HNO"_text(3(aq]) -> 2"Al"("NO"_3)_text(3(aq]) + 3"H}}_{\textrm{2 \left(g\right]}} \uparrow$

Notice the $\textcolor{red}{2} : \textcolor{p u r p \le}{6}$ mole ratio that exists between the metal and the acid. This tells you that the reaction will always consume three times as many moles of nitric acid than of aluminium metal.

Use aluminium's molar mass to determine how many moles you have in that $\text{18.0-g}$ sample

18.0color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "0.6672 moles Al"

This means that the nitric acid solution must contain

0.6672color(red)(cancel(color(black)("moles Al"))) * (color(purple)(6)color(white)(a)"moles HNO"_3)/(color(red)(2)color(red)(cancel(color(black)("moles Al")))) = "2.002 moles HNO"_3

So, molarity is defined as moles of solute, which in your case is nitric acid, per liter of solution.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} c = {n}_{\text{solute"/V_"solution}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You know that your solution has a concentration of ${\text{1.3 mol L}}^{- 1}$, and that it must contain $0.3336$ moles of solute. This means that you will have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies V_"solution" = n_"solute}} / c} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to get

V_(HNO_3) = (2.002 color(red)(cancel(color(black)("moles"))))/(1.3color(red)(cancel(color(black)("mol")))"L"^(-1)) = "1.54 L"

Rounded to two sig figs, the number of sig figs you have for the molarity of the nitric acid solution, the answer will be

V_"solution" = color(green)(|bar(ul(color(white)(a/a)"1.5 L"color(white)(a/a)|)))

SIDE NOTE It's worth mentioning that this reaction only takes place with dilute nitric acid.

Concentrated nitric acid will react with the metal and form a protective layer of metal oxide that will bring the reaction to a halt - this phenomenon is called passivation.

Moreover, the reaction will not produce hydrogen gas. Instead, the acid will be reduced to a nitrogen oxide.