Bromine is prepared commercially by the reaction: 2Br-(aq)+Cl2(aq) arrow 2Cl-(aq)+Br2(aq) Suppose we have 50.0 mL of 0.060 M solution of NaBr. What volume of 0.05 M solution of Cl2 is needed to react completely with the Br- ?

Question

Bromine is prepared commercially by the reaction: 2Br-(aq)+Cl2(aq) arrow 2Cl-(aq)+Br2(aq) Suppose we have 50.0 mL of 0.060 M solution of NaBr. What volume of 0.05 M solution of Cl2 is needed to react completely with the Br- ? (A) 30 mL (B) 40 mL (C) 20 mL (D) 60 mL

Tardigrade Admin · Accepted Answer

50 mL of 0.060 M NaBr contain NaBr =(0.060/1000)×50 mol=0.003 mol 2 mol of Br- react with Cl2 = 1 mol ∴ 0.003 mol of Br- will react with Cl2=(1/2)× 0.003 =0.0015 mol 0.05 mol Cl2 solution are present in 1000 mL of Cl2 solution ∴ 0.0015 mol of Cl2 will be present in Cl2 solution =(1000/0.05)×0.0015=30 mL

Q. Bromine is prepared commercially by the reaction: $2 B r_{(a q)}^{-} + C l_{2 (a q)} \to 2 C l_{(a q)}^{-} + B r_{2 (a q)}$ Suppose we have $50.0 m L$ of $0.060 M$ solution of $N a B r .$ What volume of $0.05 M$ solution of $C l_{2}$ is needed to react completely with the $B r^{-}$ ?

$30 m L$

$40 m L$

$20 m L$

$60 m L$

Q. Bromine is prepared commercially by the reaction: 2Br(aq)−​+Cl2(aq)​→2Cl(aq)−​+Br2(aq)​ Suppose we have 50.0mL of 0.060M solution of NaBr. What volume of 0.05M solution of Cl2​ is needed to react completely with the Br− ?

30mL

40mL

20mL

60mL

Q. Bromine is prepared commercially by the reaction: $2 B r_{(a q)}^{-} + C l_{2 (a q)} \to 2 C l_{(a q)}^{-} + B r_{2 (a q)}$ Suppose we have $50.0 m L$ of $0.060 M$ solution of $N a B r .$ What volume of $0.05 M$ solution of $C l_{2}$ is needed to react completely with the $B r^{-}$ ?

$30 m L$

$40 m L$

$20 m L$

$60 m L$