# Question #836da

Jan 30, 2016

The mass of $B {r}_{2}$ required to react with 0.9 mol of chalcone is
143.8 g.

#### Explanation:

First you want to create a chemical equation:

${C}_{15} {H}_{12} O + B {r}_{2} \to {C}_{15} {H}_{11} B r O + H B r$

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 ${C}_{15} {H}_{12} O$, 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, $B {r}_{2}$ has 0.9 mols

$n$ = $\frac{m}{M}$, which $m$ = Mass (in grams), $M$ = Molar mass and $n$ = Mols
This equation can be reformed to $m$ = $n \cdot M$

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 $B {r}_{2}$

$m$ = $0.9 \cdot 159.8$
$m$ = 143.8

Therefore the mass of $B {r}_{2}$ required to react with 0.9 mol of chalcone is 143.8 g.