session.masteringchemistry.com MasteringChemistry: Chapter Seven Homewor Masteri
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session.masteringchemistry.com MasteringChemistry: Chapter Seven Homewor MasteringChemistry Problem 7.82 Problem 7.82 Iron(III) oxide reacts with carbon monoxide to produce iron and carbon dioxide Fe20s(s) +3C0(9)+2Fels) +3C02(9) Part A What is the percent yield of iron if the reaction of 66.0 g of iron(III) oxide produces 15.6 g of iron? Express your answer with the appropriate units The percent yield of Fe Value Units Submit My Answers Give Up Part B what is the percent yeld of carbon diodetha reaction of 75.3 g of carbon monoxide produces sag g o, carbon dioxide? Express your answer with the appropriate units The percent yield of COValue Units My Answers Gie UpExplanation / Answer
Fe2O3(s) + 3CO(g) ---> 2Fe(s) + 3CO2 (g)
In this reaction 1 mole of Fe2O3 reacts with 2 moles of CO and produces 2 moles of Fe and 3 moles of CO2.
Part A: no. of moles of Fe2O3 = mass/mol.wt
= 66 g/ 159.687 g/mol
= 0.4133 moles
Therefore, theorical no. moles of Fe = no. of moles of Fe2O3 x 2
= 0.4133 x 2
= 0.8266 moles
No. of moles of Fe = = mass/mol.wt
= 15.6 g/ 55.845 g/mol
= 0.279 moles
Percent of yield of Iron = (actual no. moles/theoretical no.of moles)100
= (0.279 moles/0.8266 moles )x100
= 33.75 %
Part B:
No. of moles of CO = 75.3 g/ 28.010 g/mol
= 2.688 moles
Therefore, the theorical no. moles of CO2 = 2.688
Actual no. moles of CO2 = 88.9 g/ 44.009 g/mol
= 2.02 moles
Therefore, the percent of yield = (actual no. moles/theoretical no.of moles)x100
= (2.02 moles/2.688 moles)x100
= 75.71 %
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