Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Consider this reaction for the production of lead. 2PbO(s) + PbS(s) ---> 3Pb(s)

ID: 683695 • Letter: C

Question

Consider this reaction for the production of lead. 2PbO(s) + PbS(s) ---> 3Pb(s) +SO2(g) What is the maximum mass of lead that can beobtained by the reaction of 57.33 g PbO and 33.80 g of PbS?Please EXPLAIN your answer. Formula          Molar Mass (g/mol) Pb                        207.2 PbO                     223.2 PbS                     239.3 SO2                      64.07 Consider this reaction for the production of lead. 2PbO(s) + PbS(s) ---> 3Pb(s) +SO2(g) What is the maximum mass of lead that can beobtained by the reaction of 57.33 g PbO and 33.80 g of PbS?Please EXPLAIN your answer. Formula          Molar Mass (g/mol) Pb                        207.2 PbO                     223.2 PbS                     239.3 SO2                      64.07

Explanation / Answer

moles of PbO = mass/molar mass = 57.33/223.2 = 0.256855 mole

moles of PbS = 33.80/239.3 = 0.1412453mole

here for complete reaction of PbO, PbS required = 0.256855/2 = 0.1284275 mol so, PbS is in excess and PbO is limiting reagent.

hence, moles of Pb produced = 3/2 x moles of PbO = 3/2 x 0.256855 = 0.3852825 mole

mass of lead = moles x molar mass = 0.3852825 mole x 207.2 g/mol = 79.83 gram

maximum mass of lead = 79.83 gram

if you satisfied please rate it.. thanks...