Question #836da

1 Answer
Jan 30, 2016

The mass of #Br_2# required to react with 0.9 mol of chalcone is
143.8 g.

Explanation:

First you want to create a chemical equation:

#C_15H_12O + Br_2 -> C_15H_11BrO + HBr#

From this equation we can tell that this reaction is a double replacement reaction. This is because the bromide is more reactive than the hydrogen, and replaces the hydrogen on the benzene ring of the chalcone.

We a given 0.9 mol of #C_15H_12O#, which has a molar mass of 208.26 g/mol.

As the coeficients (the numbers in front of the compounds in the chemical equation) are the same, the 0.9 mol is applicable to use throughout the entire equation.

So, #Br_2# has 0.9 mols

#n # = #m/M#, which #m# = Mass (in grams), #M# = Molar mass and #n# = Mols
This equation can be reformed to #m# = # n*M#

Bromine has a molar mass of 79.9 mol/g. But this is one atom and must be doubled to be the molar mass of #Br_2#

#m# = #0.9*159.8#
#m# = 143.8

Therefore the mass of #Br_2# required to react with 0.9 mol of chalcone is 143.8 g.