Answer: 211.2 grams
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L)}}[/tex] .....(1)
Molarity of [tex]KBr[/tex] solution = 1.7 M
Volume of solution = 3.5 L
Putting values in equation 1, we get:
[tex]1.7M=\frac{\text{Moles of}KBr}{3.5L}\\\\\text{Moles of }KBr=\{1.7mol/L\times 3.5L}=5.95moles[/tex]
The balanced chemical reaction is:
[tex]Cl_2+2KBr\rightarrow 2KCl+Br_2[/tex]
According to stoichiometry:
2 mole of KBr requires = 1 mole of [tex]Cl_2[/tex]
5.95 moles of KBr requires = [tex]\frac{1}{2}\times 5.95=2.975[/tex] moles of [tex]Cl_2[/tex]
Mass of [tex]Cl_2=moles\times {\text {Molar mass}}=2.975mol\times 71g/mol=211.2g[/tex]
Thus 211.2 grams of chlorine gas are needed to react with 3.5 liters of a 1.7 molar potassium bromide solution
