1_05g de alchena necunoscuta aditioneaza 560 cm3 de h2
Post on 31-Dec-2015
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1,05g de alchena necunoscuta aditioneaza 560 cm3 de H2:a) formula moleculara a alchenei si formula ei de structura;b) masa de alcan obtinuta daca reactia are loc la un randament de 60%.a)Formula generala a unei alchena CnH2n
1 mol CnH2n = 14n g1 mol CnH2n+1 = 14n+2 g560 cm3 de H2=0,560LEcuatia reactiei de aditie:CnH2n + H2 → CnH2n+2
Se determina n14n g.....22,4 LCnH2n + H2 → CnH2n+2
1,05 g.......0,56L14n∙0,56 = 22,4∙1,057,84n = 23,52n=3Formula alchenei: C3H6 propena CH2=CH-CH3
Formula alcanului: C3H8 propanulb)Se determina masa de propan teoretic obtinuta (adica din calcule)1 mol C3H6 = 42 g1 mol C3H8 = 44 g42g..................44gC3H6 + H2 → C3H8
1,05g.................xx=1,1 g C3H8
Se determina masa de propan obtinut practicη=masa practica∙100/masa teoreticamasa practica = masa teoretica∙η/100 = 1,1 ∙ 60/100= 0,66 g propan
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