Question

1. Prin descompunerea a 272,2 g KCLO de puritate 90% se obține o cantitate de oxigen egala cu :
a)96g
b)67,2l
c)3*6,024*10 la puterea 23 molecule

Answer 1

MKClO₃=AK+ACl+3AO=39+35,5+3*16=122,5g/mol

MO₂=32g/mol

puritatea= masa pura/masa impura*100 => masa pura=puritatea*masa impura/100 = 90*272,2/100=244,98g KClO3

2KClO₃ ⇒ 2KCl + 3O₂

2*122,5 KClO₃........................3*32g O₂
244,98g KClO₃............................xg O₂

x=3*32*244,98/2*122,5=96g O₂

n=m/M=96/32=3 moli O₂

n=V/Vm⇒V=n*Vm=3*22,4=67,2 L O₂

1 mol O₂.........................6,023*10^23 molecule
3 moli O₂........................x=3*6,023*10^23 molecule

R: a,b,c