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

Question

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

aramutselv 2022-01-10 Answered

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

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Expert Answer

zesponderyd

Answered 2022-01-11 Author has 42 answers

Step 1

C r^{3 +} + 3 F^{-} \Rightarrow C r F_{3}

M g^{2 +} + 2 F^{-} \Rightarrow M g F_{2}

Write a balanced chemical equation
Step 2

\frac{x m o l}{1.0 L} = 1.51 M

Solve for moles using Molarity
Step 3

x = 1.51 m o l

Multiply 1.51 by 1.0 to get 1.51 moles
Step 4

49.6 g \cdot \frac{1 m o l}{109.0 g} \cdot \frac{3 m o l}{1 m o l} = 1.36 m o l

109.0 is obtained by the molar mass of CrF3

(52 + 3 (19))

Step 5

49.6 g \cdot \frac{1 m o l}{62.3 g} \cdot \frac{2 m o l}{1 m o l} = 1.59 (1 - x)

Step 6

(1.37 x + 1.59 - 1.59 x) m o l = 1.51 m o l

Step 7

x = 0.36

Step 8

0.36 \cdot 49.6 = 18 g

Multiply 0.36 by the total mass
Step 9

18 \cdot \frac{1 m o l}{(52.0 + 3 (19.0))} \cdot \frac{52.0 g}{1 m o l} = 8.6 g C r 3 +

Convert 18 grams to moles then to moles of Cr3+

This is helpful 0

Ronnie Schechter

Answered 2022-01-12 Author has 27 answers

Step 1
The molarity of the sodium fluoride (NaF) solution is 1.51 M.
The volume of the solution $= 1.00 L$ .
Now, determine the number of moles of NaF present in the solution.
Moles of $N a F = Molarity of NaF \times Volume of solution$
$= 1.51 \times 1.00 L$
$= 1.51 m o l$ .
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 $C r F_{3} and M g F_{2}$ . Assume that there are x moles of $C r^{3 +}$ ions and y moles of $M g^{2 +}$ ions in the precipitate. Since $C r F_{3}$ consists of three fluoride ions for each $C r^{+ 3}$ ion and $M g F_{2}$ consists of two fluoride ions for each $M g^{2 +}$ ion, the following equation can be written as:
$3 x + 2 y = 1.51$ (1)
The total mass of precipitate $= 49.6 g$ .
The mass of $C r F_{3} = 109.00 \frac{g}{m} o l$ .
The molar mass of $M g F_{2} = 62.31 \frac{g}{m} o l$ .
Based on this data the following equation is:
$109 x + 62.31 y = 49.6$ (2)
Step 3
Simplify and solve equations (1) and (2) to get x and y values:
$E q (1) \times 109 \Rightarrow 327 x + 218 y = 164.59$
$E q (2) \times 3 \Rightarrow 327 x + 186.93 y = 148.8$
$31.07 y = 15.79$
$y = \frac{15.79}{31.07} = 0.508 m o l$
Substitute this y value in equation (1), to get
$3 x + 2 (0.508) = 1.51$
$3 x = 0.494$
$X = 0.1646 m o l$ .
Thus $x = 0.1646 m o l and y = 0.648 m o l$ .
Step 4
The molar mass of chromium ion $= 52 \frac{g}{m} o l$ .
The number of moles of $C r^{3 +}$ ion is 0.1646 mol.
Mass of $C r^{3 +}$ ion in the original solution $= m o l e s \times$ molar mass
$= 0.1646 m o l \times 52 \frac{g}{m} o l$
$= 8.5592 g$ .
Therefore, the mass of the $C r^{3 +}$ ion is 8.5592g.