Question

Cl2 + 2KBr -> Br2 + 2KCI

How many grams of Br2 can be produced from 356 g of KBr?
A. 2.99 g Br2
B. 1.50 g Br2
C. 9.10 g Br2
D. 239 g Br2​

Answer 1

Answer: D. 239 g of
`Br_2`

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} KBr=(356g)/(119g/mol)=2.99moles`

The balanced chemical reaction is:

`Cl_2+2KBr\rightarrow Br_2+2KCl`

According to stoichiometry :

2 moles of
`KBr` produce = 1 mole of
`Br_2`

Thus 2.99 moles of
`KBr` produce =
`(1)/(2)* 2.99=1.495moles` of
`Br_2`

Mass of
`Br_2=moles* {\text {Molar mass}}=1.495moles* 159.8g/mol=239g`

Thus 239 g of
`Br_2` will be produced from 356 g of KBr.