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Dyana98
a fost răspuns

determinati masa probei de glucoza de puritate 90% stiind ca s-au obtinut 1,5 moli de argint.

Răspuns :

1mol.......................................................................2mol
C6H12O6 + 2[Ag(NH3)2]OH ---> C6H12O7 + 2Ag + 4NH3 + H2O
x=0.75mol...............................................................1.5mol

n=m/M==>m=n*M=0.75*180=135g C6H12O6 pur

p=mpur/mimpur*100==>mimpur=mpur*100/p=135*100/90=150g C6H12O6

MC6H12O6=180g/mol

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