A 2.50 g sample of bronze was dissolved in sulfuric acid. The copper in the allo

Question

A 2.50 g sample of bronze was dissolved in sulfuric acid. The copper in the alloy reacted with sulfuric acid as follows. Cu(s) + 2H_2SO_4 (aq) rightarrow CuSO_4(aq) + SO_2 (g) + 2 H_2O (I) The CuSO_4 formed in the reaction was mixed with Kl to form Cul (copper iodide). 2 CuSO_4 (aq) + 5 I^- rightarrow 2 Cul (s) + I^3 (aq) + 2 SO_4^2- (aq) The I_3^- formed in that reaction was then titrated with S_2O_ 3^2-. I_3^- (aq) + 2 S_2O_3^2- rightarrow 3 I^- (aq) + S_4O_6^2- (aq) Calculate the percentage by mass of copper in the original sample 31.5 mL of 1.00 M S_2O_3^2- was consumed in the titration.

Explanation / Answer

Moles of S2O32- needed = M * V / 1000 = 1.00 * 31.5 / 1000 = 0.0315 mol

From the last balanced equation,

2 mol S2O32- = 1 mol I3-

then, 0.0315 mol S2O32- = 1 * 0.0315 / 2 =0.01575 mol

From second balanced equation ,

1 mol of I3- is obtained from 2 mol of CuSO4

then, 0.01575 mol of I3- obtained from 2 * 0.01575 = 0.0315 mol of CusO4

From first balanced equation,

1 mol of CuSO4 is obtained from 1 mol of Cu

then, 0.0315 mol of CuSO4 is obtained from 0.0315 mol of Cu

Therefore, Mass of Cu in the original sample = 0.0315 * 63.5 = 2.00 g.

Percent by mass of Cu present in original sample = (2.00 / 2.50) * 100 = 80 %

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