Question

Two aqueous hydrogen bromide solutions containing 12.0 wt% HBr (SG = 1.1377) and 30.0 wt% HBr (SG = 1.2552) are mixed to form a 3.90 molar HBr solution (SG = 1.1234). a) What feed rate of the 30 wt% HBr solution would be required to produce 1.10x10^3 kg/hr of product? b) What is the feed ratio of litres of 12.0 wt% HBr solution to litres of 30.0 wt% HBr solution?

Answer 1

Assuming that "the product" is the required 4.27M HBr mixture, we cannot answer Part 1 until we have answered Part 2.

2.
First we need to find the wt% of HBr in the 4.27M solution. The RMM of HBr is 1+79.9=80.9, so the mass of 4.27 mol of HBr is 4.27x80.9=345g. The 4.27M solution has a SG of 1.2340, so the mass of 1.0000 litre is 1.2340kg, 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 (0.345/1.234)x100%=27.96.

Now we need to work out what volumes of 12.0 and 30.0 wt% solutions will give 1.0000 litre of 27.96 wt% HBr solution when mixed.

1.0000 litre of 4.27M (27.96 wt%) HBr solution contains 0.3450 kg of HBr and 1.2340-0.3450=0.8890 kg of water. Suppose we have to mix x litres of 12.0 and y litres of 30.0 wt% solutions to get 1.0000 litre of 4.27M solution. The total mass of HBr in these 2 volumes is 0.120*1.1377x+0.300*1.2552y=0.136524x +0.37656y kg; this must equal 0.3450 kg. Similarly, the mass of water in these 2 volumes is (1-0.120)1.1377x+(1-0.300)1.2552y=0.88*1... kg, which must equal 0.8890 kg. So we have 2 simultaneous equations which we can solve for x and y:
0.136524x+0.37656y=0.3450
1.001176x+0.87864y=0.8890.
Eliminate x by multiplying 1st eqn by 1.001176 and 2nd eqn by 0.136524 and then subtracting:
0.1366845x+0.37700283y=0.34540572
0.1366845x+0.11995545y=0.12136984
0.25704738y=0.22403588
y=0.87157426.
Now substitute for y in any eqn (I will use 1st eqn) to obtain x:
0.1366845x+0.37700283*0.87157426=0.3454...
0.1366845x=0.01681976
x=0.12305536.

ANSWER: to obtain 1 litre of 4.27M solution we must mix 0.123litre of 12.0 wt% HBr and 0.872 litre of 30.0 wt% HBr solutions. (I have rounded to 3 significant figures because that is the smallest number of sig. figs. occurring in the data.)

1.
Each litre of product (4.27M HBr solution) has a mass of 1.2340 kg, so 1.00x10^3kg of this solution has a volume of 1.00x10^3/1.2340=810.4 litres.
Each litre of product requires y=0.8716 litre of 30.0 wt% HBr solution. Therefore the feed rate for 30.0 wt% HBr is 810.4x0.8716=706 litres/hr (to 3 sig. figs.).