MH2O = 2 * 1g + 16= 18g/mol
MKClO3=1 * 39 + 1,35,5+3 * 16 g + 3 * 16g = 122,5 g/mol
n= [tex]\frac{m}{N}[/tex] = [tex]\frac{38}{16}[/tex]=2moli H2O
mO = 2 * 16 = 32g
2*122,5 3*32
2KClO3------------------> 2KlCl+3O2
x g 32 g
[tex]\frac{2*122,5}{x}[/tex] = [tex]\frac{3 *32}{32}[/tex] se taie 32 cu 32
[tex]\frac{245}{x}[/tex] = [tex]\frac{3}{1}[/tex]
x=[tex]\frac{245}{3}[/tex]
x= 81,66g KClO3