Question

Ce cantiate de glucoza in moli si grame reactioneaza cu 67,2 litrii de oxigen ? In ce cantitate de aer se gaseste acest Oxigen ? COROANAA RPD PLSS

Answer 1

C6H12O6 +9 O2 = 6CO2 +6 H2O
M C6H12O6 = 12×6+12+6×16=180g/mil
x g C6H12O6............67,2 L O2
180 g C6H12O6........9×22,4L O2
x= 60g glucoza
n= 60/180=0,33 moli glucoza
V aer ............67,2 L O2
100...................20
Vaer= 336 L