Question

Two aqueous hydrogen bromide solutions containing 1.00 wt% HBr (SG = 1.0041) and 65.0 wt% HBr (SG = 1.7613) are mixed to form a 3.19 molar HBr solution (SG = 1.1743).

Part B is correct, I just need Part A.

Answer 1

Mixed solution contains 3.19 moles of HBr/ 1 L solution

Basis :1 L solution. This contains 3.19 moles of HBr

Molecular weight of HBr= 1+80= 81

Mass of HBr= 81*3.19 gms=258.4 gm

Density of the solution = specific gravity* density of water = 1.1743*1000 kg/m3=1174.3 kg/m3

1m³ will have 1174 kg of mass

0.001m3 (1L) will have 1174*0.001=1.174 kg =1174 gm

mass of water in the solution =1174-258.4=915.6 gm of water

let x= mass of 1 wt % HBr and y= mass of 65 wt% HBr

writing mass balance on water

0.99x+0.35y= 915.6 (1)

x+0.35/0.99y= 915.6/0.99 , x +0.35y= 925 (1A)

Mass balance of HBr

0.01x+0.65y= 258.4 (2), x+65y= 25840 (2A)

Eq.2A- Eq.1A   (65-0.35)y= 25840-925

y= 385.3 kg/hr

1174 gm of solution requires 385.3 gm of 65 wt% HBr

1050 kg of solution requires 1050*385.3/1174=344.6 kg of 65 wt% HBr.

density

Feed rate of 65 wt% HBr = 344.6 kg/hr/ 1761.3kg/m3=0.195m3/hr = 195 L/hr