How many grams of 1.6% bromine water can 1.12L propylene discolor?

First of all, let’s write down the reaction of propylene with bromine water.

HBr + CH2 = CH-CH3 = CH3-CHBr-CH3

This reaction requires 1 mol of propane to decolorize 1 mol of HBr.

So we have 1.12 liters. propane, and this in moles will be 1.12: 22.4 = 0.05 (mol). Therefore, HbBr will also need 0.05 mol. And this is 81 * 0.05 = 4.05 (grams) in grams. And now we will find how much bromine water is needed. And this is 4.05 * 100: 1.6 = 253 (grams).