Calculate the solubility at 25°c of cubr in pure water and in a 0.0030m cobr2 solution. you'll find ksp data in the aleks data tab. round both of your answers to 2 significant digits. solubility in pure water: gl

the solubility at 25°c of cubr in pure water = 0.011 g/L

The solubility at 25°c of cubr  in a 0.0030m cobr2 solution = 0.00014 g/L

Explanation:

Step 1: Data given

Ksp of CuBr = 6.27 × 10^-9

Molar mass CuBr = 143.45 g/mol

Step 2: Calculate the solubility at 25°c of cubr in pure water

Ksp = [Cu+][Br-]

The initial concentration

[Cu+] = 0M

[Br-] = 0M

The concentration at the equilibrium

[Cu+] = X M

[Br-] = X M

Ksp =6.27 * 10^-9 =  X * X = X²

S = 7.9 *10^-5 mol /l  

7.9 * 10^-5 mol/L * 143.45 g/mol = 0.011 g/L

Step 3: Calculate the solubility at 25°c of cubr  in a 0.0030m cobr2 solution.

The balanced equation:

CoBr2(aq) → Co^2+(aq) + 2 Br⁻(aq)

For 1 mol CoBr2 we'll have 1 mol Co^2+ and 2 moles Br-

For 0.0030 M CoBr2 we'll have 0.0030 M Co^2+ and 0.0060 M Br-

The initial concentration

[Cu+] = 0M

[Br-] = 0.0060 M

At the equilibrium

[Cu+] = X

[Br-] = 0.0060 + X

Ksp = 6.27 * 10^-9 = X * (0.0060+X)

6.27 *10^-9 = 0.0060 X + X²

X = 1.0 * 10^-6 mol/L

1.0 * 10^-6 mol/L * 143.45 g/mol = 0.00014 g/L

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

answer; protein;