Question #836da

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Question #836da

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Answer 1 · 2016-01-30T05:16:42

The mass of $B r_{2}$ $Br_2$ required to react with 0.9 mol of chalcone is
143.8 g.

Explanation:

First you want to create a chemical equation:

$C_{15} H_{12} O + B r_{2} \to C_{15} H_{11} B r O + H B r$ $C_15H_12O + Br_2 -> C_15H_11BrO + HBr$

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 $C_{15} H_{12} O$ $C_15H_12O$ , 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, $B r_{2}$ $Br_2$ has 0.9 mols

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

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 $B r_{2}$ $Br_2$

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

Therefore the mass of $B r_{2}$ $Br_2$ required to react with 0.9 mol of chalcone is 143.8 g.

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Question #836da

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