Question

1- Consider This Equation, and complete parts A, B, C and D. 3MgO + FeO, 2FeBry 3MB 0.3 + 100gr. E 1.2 gram of MgO is reacting with 12 gram of FeBr; to give actual value of 1.2 gram of Fe2O3. Find: 1) Limiting Reagent ID) %Y

Answer 1

0 3 M,0 + a Fe Brz ---- 3 moles a moles & > 4,03 + 3 ing Br2 Imole 3 moles. © mgo : f203 Fe Brz - 8 moles . mgo febra à x 8 7 27=24 12 x = .1 = 38=1235734 4 fg 03 + la Mg Bry .: 19 mgo + 8 febry — ® mgo 3 fe brz 2 mgo : Reg og i mgo : MgBry 3x = 2X0-30 0.3 X a 10- 3 a 2= 2 xog B2 - 3X0:3 12-01 2=03 2= 0.2 0.3 mgo + 0.2 febrg - 0.1 for Oz + 03 mg Brg Fe Br = 55.6 + 3(80) fe atomic wt -35.6 = 55.6 + 240 = 295:6 9m Br atomic wt = 80 per mole. I Güven Wt of Fe Brz = 100gm. ino.of moues wt 100 mol. Wt295.6 = 0.3383 moles- mgo : febrz у тдр : fe, оз 1 тоо : то в > 0.3383 0.5094 x 3 di= 3X0-3383 x = 0.3074 3 : 3 1370.5094 : x E X =0.5094. 2:0-1691

mol, Mornic cost of Mgo = 294+16 = 40g I mole = 409 0.5090 miles € 20.2962 dion mol wt of f,02 = (55-6)+3 (16) Imole = 111•2 +48 = 159-29 0-1697 moles 0.1691 X 15 om = 26.920 gm. mol wt of mg Bre = 24+ 2(80)* i mole = 24+160 - 1849m. 0.5094 molis = 0.5094 X184 = 93.36169m mgo + Fe Bra - FegUz + mg Bry 36.9299m 93-3616gm. @ 20.2969m + loogm € 3 Mgo +8febig — 3 moles 2 moles I l'@gm id gm Reg oz + 3 mg 87 Imote l 3 moles la gm. 0.00753 mous → Wt 0 0.00753 no moles 12 -0.006 Dec molus mous actual mouw offe mgo 3 molus a moles fe Brz o molu. . 0.0406 molas yield. = 0.00753 x 159 29m = 1.199 gm. Theoritically, 0.03 moles of Mgo gures 0-01 moles of fee03 Tho-tieud : dx = 0.1218 T2=0.0609 moles. 7 .: noiol moles of mgo Meguised - 0.0609 moler Actually we have & 0.03 moles of mgo. / os Limiting Agent = mgo. D.OLX1592 = 1-5989m % Vield Actual yield X100 = 75.3768% . Theoriticalyield Xo