Question

CH3 H20 ells thel ( CH3Ctye-c Bahacce egl CH3 Cha-e-oh cu М- 88. ots 106.6 36.5 graws 3.998 0.712 mole 0.045 limiting Reagent expermend yield (8) Calc. of - Yoeld Yo xield

the amount if HCL was 13ml. fill the blank and answer the bottom and formulas also please!

thank you!

Answer 1

- CH₂ - CH₂ - Ć-on + Ha I 13 me CH₂ - CH₂ - C - a 0717 312 CHE toldo C CH₃ + H, O 3.998 go ? = 0.045 moles (theori tihend duct) 13 me of Hell is given 0.717 106.6 -0.0067 moley Cexpt product) density of Hel is required finxta r generally density of Hel is 1.18 goh / me so 13 me of Hel will weight. = 13 x 1.18 gm cento = 15:34 gm - weight pis. By gm i ny more, [mous amow so this is i 36.5 i prz ,0.420 mols alcohol is present in sexs amount the limmiting reagent expt weight of producto % yield theoritical, werden theonitical weight of product brug al 08/1 % yield de expt moles iof product theoritical méolex of product 0.0067 .045 x 100 1 0 14.88% yield.

1 mole of Hcl will produce 1 mode of H₂O - 13 me of Hell means 15.34 gm of hul = 0.42 molex 7 Ho produced ix'alto 0:42 moles molemass 3 Molecular mass 0.42 x 18 = mass i =) - mass of H₂O = 7.56 Jellycon- OM 'hed .. .CHECH, -- H2O cucu, fostho 36.5 | 106. 6. m.w gramy | 18 7.56 119.34 3.998 0.77 10.42 0.045 0.0067 0.42 mols himnitting reagenta CHijaćH2-on expt yield com) - CH 0.1179m 096.yeeld : 0.0061 0.045 x 100