Question

How many grams of chlorine gas are needed to react with 3.5 liters of a 1.7 molar

potassium bromide solution?
Cl2 + 2 KBr - 2 KCl + Br2

Answer 1

Answer: 211.2 grams

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L)}}` .....(1)

Molarity of
`KBr` solution = 1.7 M

Volume of solution = 3.5 L

Putting values in equation 1, we get:

`1.7M=\frac{\text{Moles of}KBr}{3.5L}\\\\\text{Moles of }KBr=\{1.7mol/L* 3.5L}=5.95moles`

The balanced chemical reaction is:

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

According to stoichiometry:

2 mole of KBr requires = 1 mole of
`Cl_2`

5.95 moles of KBr requires =
`(1)/(2)* 5.95=2.975` moles of
`Cl_2`

Mass of
`Cl_2=moles* {\text {Molar mass}}=2.975mol* 71g/mol=211.2g`

Thus 211.2 grams of chlorine gas are needed to react with 3.5 liters of a 1.7 molar potassium bromide solution