limiting reagent and amount of pbso4 produced 53.0g pb, 77.3g pbo2 12) Th e chem
ID: 542291 • Letter: L
Question
limiting reagent and amount of pbso4 produced 53.0g pb, 77.3g pbo2 12) Th e chemical reaction occurring during the discharge of a lead storage battery can be represented by the equation: 12) Pb(s) + PbO2 (s) +2 H2SO4 (aq) 2 PbSO4(s) + 2 H2O(1) Which is the limiting reagent and the amount of Pbs04 produced if 53.0 g of Pb, 77.3 g of PbO2, and 534 mL of 0.544 M solution of H2S04 is used? (Pb 207 g/mol; Pbo2 239 g/mol; PbSO4 303 g/mol) O. 323 mol 0.25mol 53.0 a Pb 77.3a Pb D 239 A) H2S04, 88.0 g B) H2S04, 176 g C) Pbo2, 196 g D) Pb, 77.6g E) Pb, 155g 207 534mL of 0.SH4M H2SOExplanation / Answer
According to the given data:
The no. of moles of Pb used = 53 g / 207 g mol-1 = 0.2560 mol
The no. of moles of PbO2 used = 77.3 g / 239 g mol-1 = 0.3234 mol
The no. of moles of H2SO4 used = 534*10-3 L * 0.544 mol/L = 0.2905 mol
(Note: 1 M = 1 mol/L and 1 L = 103 mL)
Now, according to the given balanced equation:
1 mole of Pb, 1 mole of PbO2 need 2 moles of H2SO4 to form 2 moles of PbSO4.
Here, 0.2905 moles of H2SO4 need 0.2905/2 = 0.1452 moles of Pb as well as 0.1452 moles of PbO2.
But here, an excess of Pb and excess of PbO2 are present, i.e. complete consumption of H2SO4 takes place.
Hence, H2SO4 is the limiting reagent and the same amount, i.e. 0.2905 mol * 303 g mol-1 ~ 88 g of PbSO4 will produce.
Therefore, the correct answer is an option A.
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