1.

Bromine is prepared commercially by the reaction : 2Br^(-)(aq)+Cl_(2)(aq)rarr 2Cl^(-)(aq)+Br_(2)(aq) Suppose we have 50.0 mL of 0.060 M solution of NaBr. What volume of 0.050 M solution of Cl_(2) is needed to react completely the Br^(-)?

Answer»


Solution :`"50 ML of 0.060 M NaBr contain NaBr"=(0.060)/(1000)XX" 50 MOL = 0.003 mol"`
`" 2 mol of "Br^(-)" REACT with Cl"_(2)= "1 mol"`
`therefore"0.03 mol of "Br^(-)" will react with "Cl_(2)=(1)/(2)xx0.03=0.0015 mol`
`"0.05 mol of "Cl_(2)" solution are present in 1000 mL of Cl"_(2)" solution"`
`therefore 0.015" mol of "Cl_(2)" will be present in "Cl_(2)" solution"=(1000)/(0.05)xx0.0015=30mL.`


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