Two aqueous hydrogen bromide solutions containing 12.0 wt% HBR (SG=1.1377) and 6

Question

Two aqueous hydrogen bromide solutions containing 12.0 wt% HBR (SG=1.1377) and 65.0 wt% HBr (SG =1.7613) are mixed to form a 3.19 molar HBr solution (SG =1.1743).

1) What feed rate of the 65.0 wt% HBr solution would be required to produce a 1.00 * 10^3 kg/hr of product?

The shark tank at an aquarium houses 5 sharks that deplete the aquarium water of a total of 75.00 g of oxygen per hour. Water from the tank is diverted through a bubbler unit that oxygenates the water using a stream of oxygen-rich air, and then is returned to the tank. At steady state, water enters the bubbler containing 0.000700 wt% O2 and exits the bubbler containing 0.001000 wt% O2. For every kilogram of the inlet water stream that enters the bubbler, 0.00500 mol of oxygen-rich air with an initial composition of 40.00 mole% O2 and 60.00 mole% N2 is fed through the bubbler. There are 8 unknowns, 8 independent equations/relations and 0 degrees of freedom.

2a) What is the flow rate of the inlet water stream?

2b) What is the mole percent of O2 in the outlet air stream?

Explanation / Answer

First we need to find the wt% of HBr in the 3.19M solution. The molecular weight of HBr is 1+79.9=80.9, so the mass of 3.19 mol of HBr is 3.19x80.9=258.07g.

The 3.19M solution has a SG of 1.1743, so the mass of 1.0000 litre is 1.1743 kg, since the mass of 1.000 litre of water is 1.0000kg. This 1.0 litre contains 0.345kg of HBr, so the wt% of HBr in the solution is (0.345/1.1743) * 100=29.37%

Now we need to find out what volumes of 12.0 and 65.0 wt% solutions will give 1.0000 litre of 29.37 wt% HBr solution when mixed.

1.0000 litre of 3.19M (29.37 wt%) HBr solution contains 0.3450 kg of HBr and 1.1743-0.3450=0.8293 kg of water.

Suppose we have to mix x litres of 12.0 and y litres of 65.0 wt% solutions to get 1.0000 litre of 3.19M solution.

The total mass of HBr in these 2 volumes is = 0.120*1.1377x+0.650*1.7613y=0.136524x + 1.1448y kg; this must equal 0.3450 kg.

Similarly, the mass of water in these 2 volumes is (1-0.120)1.1377x+(1-0.650)1.7613y=0.83*1... kg, which must equal 0.8293 kg. So we have 2 simultaneous equations which we can solve for x and y:
0.136524x + 1.1448y =0.3450
1.001176x+0.61644y=0.82930.
Eliminate x by multiplying 1st eqn by 1.001176 and 2nd eqn by 0.136524 and then subtracting:
0.1366845x+1.1500283y=0.34540572
0.1366845x+0.08413y=0.063136984
1.0658y=0.282268
y=0.264842

Each litre of product (3.19M HBr solution) has a mass of 1.1743 kg, so 1.00x10^3kg of this solution has a volume of 1.00x10^3/1.1743=851.57 litres.

Each litre of product requires y=0.264842 litre of 65.0 wt% HBr solution. Therefore the feed rate for 65.0 wt% HBr is 851.57.4 * 0.2648 =225.49 litres/hr

2A) Flow rate of inlet water stream

At steady state, water enters the bubbler containing 0.000700 wt% O2 and exits the bubbler containing 0.001000 wt% O2

Amount of O2 added in bubbler is 0.001-0.0007 = 0.0003wt% = 0.0003 g O2 per 1 g of water(or 1 ml of water)

Now, 0.0003 g O2 per 1 ml of water = 0.3 g O2 per 1 L of water.

Sharks consume 75 g O2 per hour, and we need to at least replenish 75 g O2 per hour.

Flow rate = 75 g/0.3 g = 250 L water per hour.

2B)

0.3 g O2 is added per liter which is 0.018 moles

As given in question, moles of O2 = 0.005 + 0.018 = 0.023 moles and it is 40% of total while N2 moles are 60%

Moles of N2 = 60 * 0.023 / 40 = 0.0345

Mole percent of O2 = (Moles of O2/Moles of O2 + Moles of N2) * 100

                                    = (0.023/0.023+0.0345) * 100

                                       = 0.4 * 100 = 40%

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