A solution contains Cr+3 ion and Mg+2 ion. The addition of 1.00 L of 1.55 M NaF solution is required to cause the complete precipitation of these ions as CrF3(s) and MgF2(s). The total mass of the precipitate is 49.8 g. Find the mass of Cr+3 in the original solution.

52

Why did the solution go to therapy? Because it had too many ions and needed to find some balance! But don't worry, I'll help you find the mass of Cr+3 in the original solution.

To calculate the mass of Cr+3 in the original solution, we need to find the moles of CrF3(s). Since 1 mole of CrF3(s) contains 1 mole of Cr+3 ions, we can then convert moles of CrF3(s) to moles of Cr+3 ions.

First, let's calculate the moles of CrF3(s) precipitated. We know that the molar mass of CrF3 is 63.0 g/mol, and the total mass of the precipitate is 49.8 g. Therefore, the moles of CrF3(s) precipitated is:

moles of CrF3 = mass / molar mass
moles of CrF3 = 49.8 g / 63.0 g/mol
moles of CrF3 ≈ 0.79 mol

Now, since 1 mole of CrF3(s) contains 1 mole of Cr+3 ions, the moles of Cr+3 ions is also approximately 0.79 mol.

Finally, we can calculate the mass of Cr+3 in the original solution using the moles of Cr+3 and the molar mass of Cr+3 (51.99 g/mol):

mass of Cr+3 = moles of Cr+3 * molar mass
mass of Cr+3 = 0.79 mol * 51.99 g/mol
mass of Cr+3 ≈ 41 g

Therefore, the mass of Cr+3 in the original solution is approximately 41 grams.

To find the mass of Cr+3 in the original solution, we first need to determine the number of moles of Cr+3 ions in the precipitate. From the balanced equation:

2CrF3(aq) + 3MgF2(aq) -> Cr2F6(s) + 3MgF2(s)

We can see that the stoichiometry of CrF3 to Cr+3 is 2:1. This means for every 2 moles of CrF3 precipitated, we have 1 mole of Cr+3 ions.

First, let's find the number of moles of CrF3 precipitated:

Molar mass of CrF3 = atomic mass of Cr + 3*(atomic mass of F)
= 52 + 3*(19)
= 129 g/mol

Number of moles of CrF3 precipitated = mass of precipitate / molar mass of CrF3
= 49.8 g / 129 g/mol
= 0.386 mol

Since the stoichiometry of CrF3 to Cr+3 is 2:1, the number of moles of Cr+3 ions is half the number of moles of CrF3:

Number of moles of Cr+3 = 0.386 mol / 2
= 0.193 mol

Finally, let's calculate the mass of Cr+3 in the original solution:

Mass of Cr+3 in the original solution = number of moles of Cr+3 ions * molar mass of Cr+3
= 0.193 mol * molar mass of Cr+3

To find the molar mass of Cr+3, we need the atomic mass of Cr:

Atomic mass of Cr = 52 g/mol

So,

Mass of Cr+3 in the original solution = 0.193 mol * 52 g/mol
= 10.036 g

Therefore, the mass of Cr+3 in the original solution is approximately 10.036 grams.

SO WHATS THE ANSWER??

The answer is 9.95 g

Let X = mass Cr^+3

Let Y = mass Mg^+2
mm stands for molar mass.
=====================
two equations.
moles Cr + mols Mg = total moles
mass CrF3 + mass MgF2 = total mass.
===============================
X(1/mmCr) + Y(1/mmMg) = 1.55
X(mmCrF3/mmCr) + Y(mmMgF2/mmMg) = 4.98
Solve the two equation for X and Y..
Check my work.