a)\(4Na+O2-->2Na2O\)
b)\(n_{Na}=\frac{55,2}{23}=2,4\left(mol\right)\)
\(n_{O2}=\frac{1}{4}n_{Na}=0,6\left(mol\right)\)
\(V_{O2}=0,6.22,4=13,44\left(l\right)\)
c)\(2KCLO3--.2KCl+3O2\)
\(n_{KClO3}=\frac{2}{3}n_{O2}=0,4\left(mol\right)\)
\(m_{KClO3}=0,4.122,5=49\left(g\right)\)
4Na+O2-->2Na2O
2,4--0,6-- mol
nNa=55,2\23=2,4 mol
=>VO2=0,6.22,4=13,44 l
2KClO3--->2KCl+3O2
0,4--------------------0,6 mol
=>mKClO3=0,4.122,5=49g