# A solution contains Cr^{3+}ions and Mg^{2+}Mg ions. The addition of

A solution contains $C{r}^{3+}$ ions and $M{g}^{2+}Mg$ ions. The addition of 1.00 L of 1.51 M NaF solution causes the complete precipitati on of these ions as $Cr{F}_{3}\left(s\right)$ and $Mg{F}_{2}\left(s\right)$. The total mass of the precipitate is 49.6g. Find the mass of $C{r}^{3+}$ in the original solution.

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zesponderyd
Step 1
$C{r}^{3+}+3{F}^{-}⇒Cr{F}_{3}$
$M{g}^{2+}+2{F}^{-}⇒Mg{F}_{2}$
Write a balanced chemical equation
Step 2

Solve for moles using Molarity
Step 3
$x=1.51mol$
Multiply 1.51 by 1.0 to get 1.51 moles
Step 4

109.0 is obtained by the molar mass of CrF3 $\left(52+3\left(19\right)\right)$
Step 5

Step 6
$\left(1.37x+1.59-1.59x\right)mol=1.51mol$
Step 7
$x=0.36$
Step 8
$0.36\cdot 49.6=18g$
Multiply 0.36 by the total mass
Step 9

Convert 18 grams to moles then to moles of Cr3+
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Ronnie Schechter

Step 1
The molarity of the sodium fluoride (NaF) solution is 1.51 M.
The volume of the solution $=1.00L$.
Now, determine the number of moles of NaF present in the solution.
Moles of $NaF=\text{Molarity of NaF}×\text{Volume of solution}$
$=1.51×1.00L$
$=1.51mol$.
1 mol of NaF gives 1 mole of $N{a}^{+}$ ion and 1 mol of ${F}^{-}$ ion. This indicates that there are 1.51 moles of F- ions present in the precipitate.
Step 2
According to the given data, 1.55 moles of ${F}^{-}$ ions are required to complete the precipitation of . Assume that there are x moles of $C{r}^{3+}$ ions and y moles of $M{g}^{2+}$ ions in the precipitate. Since $Cr{F}_{3}$ consists of three fluoride ions for each $C{r}^{+3}$ ion and $Mg{F}_{2}$ consists of two fluoride ions for each $M{g}^{2+}$ ion, the following equation can be written as:
$3x+2y=1.51$ (1)
The total mass of precipitate $=49.6g$.
The mass of $Cr{F}_{3}=109.00\frac{g}{m}ol$.
The molar mass of $Mg{F}_{2}=62.31\frac{g}{m}ol$.
Based on this data the following equation is:
$109x+62.31y=49.6$ (2)
Step 3
Simplify and solve equations (1) and (2) to get x and y values:
$Eq\left(1\right)×109⇒327x+218y=164.59$
$Eq\left(2\right)×3⇒327x+186.93y=148.8$
$31.07y=15.79$
$y=\frac{15.79}{31.07}=0.508mol$
Substitute this y value in equation (1), to get
$3x+2\left(0.508\right)=1.51$
$3x=0.494$
$X=0.1646mol$.
Thus .
Step 4
The molar mass of chromium ion $=52\frac{g}{m}ol$.
The number of moles of $C{r}^{3+}$ ion is 0.1646 mol.
Mass of $C{r}^{3+}$ ion in the original solution $=moles×$ molar mass
$=0.1646mol×52\frac{g}{m}ol$
$=8.5592g$.
Therefore, the mass of the $C{r}^{3+}$ ion is 8.5592g.