Two aqueous hydrogen bromide solutions containing 1.00 wt% HBr (SG = 1.0041) and

Question

Two aqueous hydrogen bromide solutions containing 1.00 wt% HBr (SG = 1.0041) and 30.0 wt% HBr (SG = 1.2552) are mixed to form a 3.90 molar HBr solution (SG = 1.1234).

1.What feed rate of the 30.0 wt% HBr solution would be required to produce 1050 kg/hr of product?

2.What is the feed ratio of liters of 1.00 wt% HBr solution to liters of 30.0 wt% HBr solution?

m1=68.79,m2=981.21

my kg/hr solution (unknown) V1 L/hr solution (unknown) x1 = 1.00 wt% HBr X2 = 99.00 wt% water m3 = 1050 kg/hr solution V3 L/hr solution (unknown) X5 = 28.1 wt % HBr XG = 71.9 wt% water Mixer m2 kg/hr solution (unknown) V2 L/hr solution (unknown) X3-30.0 wt% HBr X4=70.0 wt% water

Explanation / Answer

1) Mass of one litre of the solution = 1.0041

Mass of HBr in one litre of the solution = 0.01*1.0041

= 0.010041 kg

so there is 0.010041 in one litre of the solution.

for the other feed (m2),
Mass of one litre of the solution = 1.2552 kg

Mass of HBr in one litre of the solution = 0.3*1.2552
= 0.37656 kg

so there is 0.37656 HBr in one litre of the solution.

Therefore,

balancing the masses,

m1+m2=1050

balancing the volume,

m1/0.010041 + m2/0.37656 = 1050/(0.281*1.1234)

solving, we get m1 = 68.79 kg/hr

m2= 981.21 kg/hr

2) feed ratio = V1/V3

= 68.79*0.281*1.1234/(1050*0.01*1.0041)

=2.059

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