Question

1) Interpreteaza chimic si rezolva matematic:

Informatie: Formula generala a zaharidelor este CnH2mOm. Pentru glucoza , n=m=6

A) Determina formula moleculara a glucozei
B) Calculeaza compozitia procentuala a glucozei
C) Cate grame de hidrogen se gasesc in trei moli de glucoza?
D) Ce concentratie are solutia obtinuta la dizolvarea a 5 moli de glucoza intr-un litru de apa (P H20=1g/cm³)

2) Masa hidrogenului cuprins in 10 molide H2SO4 se regaseste in x g de HCl si in ymoli de HCl.

Determina valorile lui x si y

Va rog!

Answer 1

1. a) Pentru n=m=6 => C6H12O6 - glucoza

b) MC6H12O6=12*6+12+6*16=180g/mol

mC=12*6=72g    ⇒ %C=mC/M*100=72/180*100=40%

mH=12*1=12g    ⇒ %H=mH/M*100=12/180*100=6,666%

mO=16*6=96g    ⇒ %O=mO/M*100=96/180*100=53,33%

c) Daca 1 mol C6H12O6 se gasesc 12g H
Atunci 3 moli ....................................x=12*3=36g H

d) n=md/M=>md=n*M=5*180=900g C6H12O6

roH2O=1g/cm3    ⇒ mH2O=1000g

ms=md+mH2O=900+1000=1900g sol. C6H12O6

Cp=md/ms*100=900/1900*100=47,368%

2) MH2SO4 = 2+32+64=98g/mol

1mol H2SO4.......................2gH
10 mol..............................20g H

MHCl=1+35,5g=35,5g/mol

1mol HCl.......................1gH
x=20mol........................20gH

n=m/M=>m=n*M=20*36,5=730g HCl (y)