Question

A chemist weighed out 1.654 g of a mixture containing unknown amounts of iron (Fe) and lead (Pb) and placed the sample in a 0.500 L flask containing O2 (g) at 24.0 C and 846 torr. After heating, the reaction to form Fe2O3(s) and PbO2(s) was completed, the vessel was cooled to 24.0 C and the pressure of O2 (g) remaining was 452 torr. Calculate the mass percentage of lead in the mixture. Molar masses (g/mol) : PbO2: 239.20; Fe2O3: 159.69; Pb: 207.20; Fe: 55.85

Answer 1

we first find Moles of O2

P = 846 torr = 846/760 atm = 1.11316 atm , V = 0.5 L , T = 24C = 24+273 = 297 K

we use PV = nRT equation

1.11316 x 0.5 = nO2 x 0.08206 x 297

nO2 = 0.022837

after reaction P remaining was 452 torr = 452/760 = 0.594737 atm

now ( 0.594737 x 0.5) = nO2 x 0.08206 x 297

nO2 left = 0.0122

O2 moles reacted = ( 0.022837 -0.0122) = 0.0106357

Fe mass + Pb mass = 1.654 , we have formula mass = moles x molar mass

hence 55.85 x Fe moles + 207.2 Pb moles = 1.654 .................(1)

now   2Fe + 1.5O2 ---> Fe2O3   , hence O2 moles reacted with Fe = ( 1.5/2) Fe moles = 0.75 Fe moles

Pb + O2 ---> PbO2 , O2 moles reacted with Pb = Pb moles   = 1 Pb moles

now 0.75 Fe moles + 1Pb moles = O2 total moles reacted = 0.0106357 .....(2)

now we solve (1) (2) , substituting (2) in (1) we get

Fe moles = 0.005522 , Pb moles= 0.006494

Pb mass = moles x amolar mas sof Pb = 0.006494 x 207.2 = 1.3456 g

mass % of Pb in ixture = ( 100 x Pb mass/sample mass) = ( 100 x 1.3456 /1.654) = 81.35 %