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chem 數3題20分20分!!!!!!!!!!!!!!!

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initial mass of Cu strip =4.36gfinal mass of Cu strip=2.21ga)write chemical equation b)find mass of Ag formed in reactionreactive actomic mass :Cu=63.5 Ag=108MnO2+4HCL(aq)→MnCl2(aq)+Cl2(g)+2H2O(l)217g of MnO2 treated with HCl contain 274g of HCl a) limiting agentb)mass of Cl producedc)mass of excess... 顯示更多 initial mass of Cu strip =4.36g final mass of Cu strip=2.21g a)write chemical equation b)find mass of Ag formed in reaction reactive actomic mass :Cu=63.5 Ag=108 MnO2+4HCL(aq)→MnCl2(aq)+Cl2(g)+2H2O(l) 217g of MnO2 treated with HCl contain 274g of HCl a) limiting agent b)mass of Cl produced c)mass of excess reagent left R.A.mass :H=1 O=16 CL=35.5 Mn=55 ZnO+C→Zn+CO2 48.6g of ZnO reduced with C a)find mass of Zn b)mass of C R.A mass:C=12 O=16 Zn=65

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1(a) Cu(s) + 2Ag+(aq)→Cu2+(aq) + 2Ag(s) 1(b) No. of moles of Cu used = (4.36-2.21)/63.5 = 0.03386 mole No. of moles of Ag formed = 2 * No. of moles of Cu used = 2 * 0.03386 mole = 0.06772 mole Mass of Ag formed = 0.06772 * 108 = 7.3134 g 2(a)HCl(bcos 274g HCl(7.50685 mole) cannot totally react with 217g of MnO2(2.494253 mole)) 2(b)no. of mole of HCl used = 274/(1+35.5) = 7.50685 mole no. of mole of Cl2 formed = 0.25 * no. of mole of HCl used = 0.25* 7.50685 = 1.8767mole 2(c)no. of mole MnO2 used = 0.25 * no. of mole of HCl used = 0.25* 7.50685 = 1.8767mole mass of MnO2 left = 217 - 1.8767(55 + 16*2) = 53.7260 g Balanced equation: 2ZnO+C→2Zn+CO2 3(a)mass of Zn = 48.6/(65+16)*65 = 39g 3(b)no. of mole of ZnO used = 48.6/(65+16) = 0.6 mole no. of mole of C used = 0.5*no. of mole of ZnO used = 0.5*0.6 = 0.3mole mass of C used = 0.3*12 = 3.6 g

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