How many grams of HBr would exactly react with 2g of propene?

3.84 g HBr

Explanation:

A reaction between Propene and HBr is given by the equation;

HBr + C₃H₆ → C₃H₇Br

Note; this is an addition reaction where hydrogen halides are added to an alkene.

We are given 2 g of propene

Required to determine the mass of HBr used;

We first determine the number of moles of propene used

Moles = Mass ÷ molar mass

Molar mass of propene = 42.08 g/mol

Therefore;

Moles of propene = 2 g ÷ 42.08 g/mol

                               = 0.0475 moles

Then, from the equation, 1 mole of HBr reacts with 1 mole of propene

Thus;

Moles of HBr = Moles of propene

Moles of HBr = 0.0475 moles

Lastly, we determine the mass of HBr

Mass = Moles × Molar mass

Molar mass of HBr = 80.91 g/mol

Therefore;

Mass of HBr = 0.0475 moles × 80.91 g/mol

                     = 3.84 g

indium, and thallium have three electrons i think