Answer:
78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene
Explanation:
[tex]C_3H_6+HBr\rightarrow C_3H_7Br[/tex]
Moles of propene = [tex]\frac{25.5 g}{39 g/mol}=0.6538 mol[/tex]
According to reaction, 1 mole of propene gives 1 mole of propane.
Then 0.6538 moles of bromo-propane will give:
[tex] 0.6538 mol\times 120 g/mol=78.46 g[/tex]
78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene.
