Question

A 10–mole % fluoboric acid aqueous solution is added to-60 mole% fluoboric acid aqueous solution to obtain 7,200-lb-mole of 60-w%fiuoboric acid aqueous solution.calculate the amounts(g) of 10-mole% fluoboric acid aqueous solution and the 60- mole% fluoboric acid aqueous solution to obtain 60-w% fluoboric acid aqueous solution, using the chemical Engineering handbook. Find the mean molecular weight of the 60-w% fluoboric acid aqueous solution.

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Answer 1

We need 7,200-lb-mole of 60-w%fluoboric acid aqueous solution.

Hence moles of fluoboric acid(HBF4) and water =  7,200-lb-mole = 7200 lb-mol x ( 453.59 mol / 1 lb-mol)

= 3265848 mol (HBF4 + H2O)

Let the moles of HBF4 in the final mixture be 'y' mol

Hence moles of H2O = (3265848 - y) mol

Molecular mass of HBF4 = 87.8 g/mol

Mass of HBF4 =  y mol x 87.8 g/mol

Mass of H2O = (3265848 - y) mol x 18.0 g/mol

Hence total mass of 7,200-lb-mole of 60-w%fiuoboric acid aqueous solution

= y mol x 87.8 g/mol + (3265848 - y) mol x 18.0 g/mol

= (69.8y + 58785264) g

Since 60-w%fiuoboric acid aqueous solution is prepared, hence

87.8y g / (69.8y + 58785264) g = 0.6

=> 87.8y g =  (69.8y + 58785264) x 0.6

=> y = 768100 mol

Hence moles of HBF4 in the aqueous solution = 768100 mol

and moles of water in the aqueous solution = (3265848 - y) = 2497748 mol H2O

Now we need to add 10–mole % HBF4 aqueous solution and 60 mole% HBF4 aqueous solution to obtain the above solution.

Suppose the moles of HBF4 in the 10–mole % HBF4 aqueous solution added to prepare the final solution be 'z' mol.

Hence moles of H2O in  10–mole % HBF4 aqueous solution = 9z mol

Now for 60–mole % HBF4 aqueous solution, moles of HBF4 = (768100 - z) mol

moles of H2O = (2497748 - 9z) mol and

(768100 - z) / [(768100 - z) + (2497748 - 9z)] = 0.6

=> z = 238282 mol HBF4

Hence

For 10–mole % HBF4 aqueous solution:

Moles of HBF4 = z = 238282 mol HBF4

Mass of HBF4 = 238282 mol x 87.8 g/mol = 20921160 g

Moles of H2O = 9z = 9x238282 mol = 2144538 mol H2O

Mass of H2O = 2144538 mol x 18.0 g/mol = 38601684 g H2O

Hence total mass of 10–mole % fluoboric acid aqueous solution added = 20921160 g+38601684 g

= 59522844 g (answer)

For 60–mole % HBF4 aqueous solution:

Moles of HBF4 = (768100 - z)  = (768100 - 238282) mol = 529818 mol HBF4

Mass of HBF4 = 529818 mol x 87.8 g/mol = 46518020 g

Moles of H2O = (2497748 - 9z)  = (2497748 - 9x238282) mol = 353210 mol H2O

Mass of H2O = 353210 mol x 18.0 g/mol = 6357780 g H2O

Hence total mass of 60–mole % fluoboric acid aqueous solution added = 529818 g+6357780 g

= 6887598 g (answer)