4Na+ O2-> 2Na2O
nO2= 2,24/22,4=0,1mol
nNa=4,6/23=0,2 mol
Tỉ lệ: 0,2/4< 0,1/1=> Na td hết, O2 dư
nNa2O= (0,2.2)/4=0,1 mol
m Na2O= 0,1.(23.2+16)=6,2g
2.
Zn+S-to>ZnS
tổng chất pư=2,6+1,92=4,52 g
=>%mZn=2,6\4,52.100=57,5%
=>%mS=1,92\4,52.100=42,5%
4Na+O2-to->2Na2O
nNa=4,6\23=0,2 mol
nO2=2,24\22,4=0,1 mol
=>O2 dư
mNa2O=0,05.62=3,1g