## Consider 2 crystalohydrates $N i \left(N O 3\right) 2 \setminus \cdot n H 2 O$ and $F e \left(N O 3\right) 3 \setminus \cdot 1.5 n H 2 O$.These crystalohydrates are in a mixture which has the mass of 98,6g and they are in a molar ratio of 2:1 .This mixture is disolved in 201,4g H2O resulting in a solution in which the masic concentration (In percentages) of Ni(NO3)2 is 12,2%. 1)What is the masic concentration of Fe(NO3)3? 2)What are the formulas of the 2 crystalohydrates?

Jun 6, 2017

Here's what I got.

#### Explanation:

For part (a), start by using the solution's percent concentration by mass to determine how many grams of anhydrous nickel(II) nitrate it contains.

You know that the mixture has a total mass of $\text{98.6 g}$ and that you're adding it to $\text{201.4 g}$ of water. This means that the total mass of the solution will be

${m}_{\text{solution" = "98.6 g" + "201.4 g}}$

${m}_{\text{solution" = "300.0 g}}$

As you know, the solution's percent concentration by mass is calculated by dividing the mass of the solute by the total mass of the solution and multiplying the result by 100%.

This means that you have

("mass of Ni"("NO"_ 3)_ 2)/"300.0 g" * 100color(red)(cancel(color(black)(%))) = 12.2color(red)(cancel(color(black)(%)))

which gets you

$\text{mass of Ni"("NO"_ 3)_ 2 = (12.2 * "300 g")/100 = "36.6 g}$

Next, use the molar mass of nickel(II) nitrate to calculate the number of moles of this anhydrous salt present in the initial mixture and, consequently, in the solution

36.6 color(red)(cancel(color(black)("g"))) * ("1 mole Ni"("NO"_3)_2)/(182.703color(red)(cancel(color(black)("g")))) = "0.2003 moles Ni"("NO"_3)_2

Now, you know that every mole of nickel(II) nitrate hydrate contains

• one mole of anhydrous nickel(II) nitrate, 1 xx "Ni"("NO"_3)_2
• $n$ moles of water, $n \times \text{H"_2"O}$

This means that the initial mixture contained $0.2003$ moles of nickel(II) nitrate hydrate.

You also know that the two hydrates are present in the initial mixture in a $2 : 1$ mole ratio.

$\left(\text{moles of Ni"("NO"_3)_2 * n"H"_2"O")/("moles of Fe"("NO"_3)_3 * 1.5n"H"_2"O}\right) = \frac{2}{1}$

This means that you have

"0.2003 moles"/("moles of Fe"("NO"_3)_3 * 1.5n"H"_2"O") = 2/1

which gets you

$\text{moles of Fe"("NO"_3)_3 * 1.5n"H"_2"O" = "0.10015 moles}$

Once again, you know that every mole of iron(III) nitrate hydrate contains

• one mole of anhydrous iron(III) nitrate, 1 xx "Fe"("NO"_3)_3
• $1.5 n$ moles of water, $1.5 n \times \text{H"_2"O}$

This means that the solution will contain $0.10015$ moles of anhydrous iron(III) nitrate. Use the compound's molar mass to convert this to grams

0.10015 color(red)(cancel(color(black)("moles Fe"("NO"_3)_3))) * "241.86 g"/(1color(red)(cancel(color(black)("mole Fe"("NO"_3)_3)))) = "24.22 g"

The percent concentration of iron(III) nitrate in the solution will thus be

$\left(24.22 \textcolor{red}{\cancel{{\textcolor{b l a c k}{{\text{g"))))/(300color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)("8.07% Fe"("NO}}_{3}}}_{3}}}\right)$

The answer is rounded to three sig figs.

For part (b), use the mass of the initial mixture and the masses of the two anhydrous compounds to calculate the total mass of water of hydration present in the initial mixture.

"mass of water of hydration" = "98.6 g" - ("36.6 g" + "24.22 g")

$\text{mass of water of hydration = 37.78 g}$

Next, use the molar mass of water to convert the mass to moles

37.78 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "2.10 moles H"_2"O"

Now, you know from the chemical formula of the nickel(II) nitrate hydrate that it contains

$n \cdot \text{0.2003 moles} = \left(0.2003 \cdot n\right)$ $\text{moles H"_2"O}$

Similarly, the chemical formula of the iron(III) nitrate hydrate tells you that it contains

$1.5 n \cdot \text{0.10015 moles} = \left(0.150 n\right)$ $\text{moles H"_2"O}$

You can thus say that you have

$\left(0.2003 n\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles H"_2"O"))) + (0.150n)color(red)(cancel(color(black)("moles H"_2"O"))) = 2.10color(red)(cancel(color(black)("moles H"_2"O}}}}$

Solve for $n$ to find

$0.350 n = 2.10 \implies n = \frac{2.10}{0.350} = 6$

Therefore, the two hydrates are

• $\text{Ni"("NO"_3)_2 * 6"H"_2"O} \to$ nickel(II) nitrate hexahydrate

• $\text{Fe"("NO"_3)_3 * 9"H"_2"O} \to$ iron(III) nitrate nonahydrate