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

aramutselv 2022-01-10 Answered

A solution contains Cr3+ ions and Mg2+Mg ions. The addition of 1.00 L of 1.51 M NaF solution causes the complete precipitati on of these ions as CrF3(s) and MgF2(s). The total mass of the precipitate is 49.6g. Find the mass of Cr3+ in the original solution.

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Expert Answer

zesponderyd
Answered 2022-01-11 Author has 42 answers
Step 1
Cr3++3FCrF3
Mg2++2FMgF2
Write a balanced chemical equation
Step 2
x mol1.0L=1.51M
Solve for moles using Molarity
Step 3
x=1.51mol
Multiply 1.51 by 1.0 to get 1.51 moles
Step 4
49.6g1 mol109.0g3 mol1 mol=1.36mol
109.0 is obtained by the molar mass of CrF3 (52+3(19))
Step 5
49.6g1 mol62.3g2 mol1 mol=1.59(1x)
Step 6
(1.37x+1.591.59x)mol=1.51mol
Step 7
x=0.36
Step 8
0.3649.6=18g
Multiply 0.36 by the total mass
Step 9
181 mol(52.0+3(19.0))52.0g1 mol=8.6g Cr3+
Convert 18 grams to moles then to moles of Cr3+
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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 =1.00L.
Now, determine the number of moles of NaF present in the solution.
Moles of NaF=Molarity of NaF×Volume of solution
=1.51×1.00L
=1.51mol.
1 mol of NaF gives 1 mole of Na+ 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 CrF3 and MgF2. Assume that there are x moles of Cr3+ ions and y moles of Mg2+ ions in the precipitate. Since CrF3 consists of three fluoride ions for each Cr+3 ion and MgF2 consists of two fluoride ions for each Mg2+ ion, the following equation can be written as:
3x+2y=1.51 (1)
The total mass of precipitate =49.6g.
The mass of CrF3=109.00gmol.
The molar mass of MgF2=62.31gmol.
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(1)×109327x+218y=164.59
Eq(2)×3327x+186.93y=148.8
31.07y=15.79
y=15.7931.07=0.508mol
Substitute this y value in equation (1), to get
3x+2(0.508)=1.51
3x=0.494
X=0.1646mol.
Thus x=0.1646mol and y=0.648mol.
Step 4
The molar mass of chromium ion =52gmol.
The number of moles of Cr3+ ion is 0.1646 mol.
Mass of Cr3+ ion in the original solution =moles× molar mass
=0.1646mol×52gmol
=8.5592g.
Therefore, the mass of the Cr3+ ion is 8.5592g.

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