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What is the amount of CuSO4 present in a given sample using coulometry if a constant current of 50 mA is passed through the sample for 30 minutes, and 30.6 mA-h of charge is consumed in the process?

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To determine the amount of CuSO4 present in the sample, we can follow these steps:1. Convert the given current and time to coulombs  charge .2. Calculate the moles of electrons transferred.3. Determine the moles of CuSO4.4. Calculate the mass of CuSO4.Step 1: Convert current and time to coulombsCurrent  I  = 50 mA = 50 x 10^-3  ATime  t  = 30 minutes = 30 x 60 secondsCharge  Q  = I x tQ =  50 x 10^-3  A  x  30 x 60 s Q = 90,000 x 10^-3  CQ = 90 CStep 2: Calculate the moles of electrons transferredGiven that 30.6 mA-h of charge is consumed, we need to convert this to coulombs as well.Charge consumed = 30.6 mA-h x  1000 mA/A  x  3600 s/h Charge consumed = 30.6 A-sMoles of electrons  n  = Charge consumed / Faraday's constantn = 30.6 A-s / 96485 C/moln  0.000317 molStep 3: Determine the moles of CuSO4The balanced equation for the reduction of CuSO4 is:CuSO4 + 2e-  Cu + SO4^2-From the stoichiometry, 1 mole of CuSO4 requires 2 moles of electrons. Therefore,Moles of CuSO4 = Moles of electrons / 2Moles of CuSO4  0.000317 mol / 2Moles of CuSO4  0.0001585 molStep 4: Calculate the mass of CuSO4Molar mass of CuSO4 = 63.5  Cu  + 32  S  + 4 x 16  O  = 159.5 g/molMass of CuSO4 = Moles of CuSO4 x Molar mass of CuSO4Mass of CuSO4  0.0001585 mol x 159.5 g/molMass of CuSO4  0.0253 gTherefore, the amount of CuSO4 present in the given sample is approximately 0.0253 grams.

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