Bromine is prepared commercially by the reaction: Br−(aq)+Cl2(aq)→Cl−(aq)+Br2(aq) suppose we have 50.0 mL of 0.060 M solution of NaBr. What volume of 0.050 M solution of Cl2 is needed to react completely with the Br− ?
A
50 mL
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B
10 mL
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C
25 mL
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D
30 mL
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Solution
The correct option is D 30 mL 50 mL of 0.060 M NaBr contain NaBr =0.0601000×50=0.003 mol 2 mol of Br− react with 1 mol of Cl2 because the balanced equation is 2Br−+Cl2→2Cl−+Br2 ∴0.003 mol Br− will react with =12×0.003 = 0.0015 mol Cl2 Given, 0.05 mol of Cl2 solution is present in 1000 mL of Cl2 solution ∴ volume of Cl2=10000.05×0.0015=30 mL