che owLv2] Online teaching x Astudent is given a sampG Search Q&A1Cheggcomx; gen

Question

che owLv2] Online teaching x Astudent is given a sampG Search Q&A1Cheggcomx; genow.com/ilrn/takeAssignment/takeCovalentActivity.do?locator-assignment-take&takeAssignmentSessionLocator-assign Pure copper may be produced by the reaction of copper() sulfide with oxygen gas as follows Cu2S(s) + O2(g) 2Cu(s) + SO2(g) If the reaction of 1.500 kg of copper) sulfide with excess oxygen prodaces 1 100 kg of coper metal that is the perces yizid A. 73.3 % B. 39.9 % C. 91.8% OD, 184 % E. 367 % 1 Rorm attempt remaining Try Another Version Sebmit Answer

Explanation / Answer

We calculate theoretical yield and by using it we get percent yield.

Theoreotical yield is calculated by using reaction stoichiometry. To calculate theoretical yield limiting reactant is important since it limits the formation of product. Here in this reaction O2 is in excess so Cu2S is limiting reactant.

Calculation of mol of Cu2S

Mole of Cu2S = 1.500 kg * 1000 g * 1 mol Cu2S / Molar mass of Cu2O

= 1.500 kg * 1000 g * 1 mol Cu2S / 159.158 g

= 9.4246 mol Cu2S

Calculation of moles of Cu(s) :

Mole of Cu(s) = moles of Cu2S * 2 mol Cu(s) / 1 mol Cu2S

= 18.85 mol Cu(s)

Lets calculate mass of Cu(s)

Mass of Cu(s) = 18.85 mol Cu(s) * Molar mass of Cu

                        = 18.85 mol Cu(s) * 63.546 g per mol

                        =1197.79 g Cu(s)

Mass of Cu in kg

                        = 1197.79 g Cu(s) * 1 kg / 1000 g

                        = 1.198 kg Cu(s)

Calculation of percent yield:

Percent yield = (Actual yield / theoretical yield ) X 100

=( 1.100 kg / 1.198 kg ) X 100

=91.8 %

So the answer is 91.8 %

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.