PTHH: 4Na + O2 --to--> 2Na2O
a) \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
Theo PTHH: \(n_{O_2}=\dfrac{1}{4}n_{Na}=\dfrac{1}{4}.0,2=0,05\left(mol\right)\)
\(V_{O_2\left(đktc\right)}=0,05.22,4=1,12\left(g\right)\)
b) Theo PTHH: \(n_{Na_2O}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(m_{Na_2O}=0,1.62=6,2\left(g\right)\)
a)ta có PTHH: 4Na +O2-->2Na2O
nNa=4,6:23=0,2mol
theo PTHH 4mol Na cần 1mol Oxi
0,2 mol Na cần 0,05mol O2
VO2=0,05.22,4=1,12(l)
b) theo PTHH 1mol Oxi thành 2 mol Na2O
0,05mol Oxi thành 0,1 mol Na2O
m Na2O=0,1.62=6,2g
PTHH:4Na+O2----->2Na2O
a.\(n_{Na}=\dfrac{m_{Na}}{M_{Na}}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
Theo PTHH:\(n_{O_2}=\dfrac{1}{4}n_{Na}=\dfrac{1}{4}.0,2=0,05\left(mol\right)\)
\(V_{O_2}=n_{O_2}.22,4=0,05.22,4=1,12\left(l\right)\)
b.Theo PTHH:\(n_{Na_2O}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(m_{Na_2O}=n_{Na_2O}.M_{Na_2O}=0,1.62=6,2\left(g\right)\)
Chúc bạn học tốt
PT: 4Na + O\(_2\)\(\underrightarrow{t}\)2Na\(_2\)O
a) n\(_{Na}\)=\(\dfrac{4,6}{23}\)=0,2(mol)
Theo PT ta có: n\(_{O_2}\)=\(\dfrac{1}{4}\)n\(_{Na}\)=\(\dfrac{1}{4}\).0,2=0,05(mol)\(\Rightarrow\)V\(_{O_2}\)=22,4.0,05=1,12(l)
b)Theo PT ta có: n\(_{Na_2O}\)=\(\dfrac{1}{2}\)n\(_{Na}\)=\(\dfrac{1}{2}\). 0,05=0,025(mol)
\(\Rightarrow\)m\(_{Na_2O}\)=0,025.62=1,55(g)
nNa =\(\dfrac{m}{M}\) =\(\dfrac{4,6}{23}=0,2\) (mol)
ta có pt :
Na + O2 → NaO2
1mol 1mol
0,2 mol
nO2=0,2.1:1=0,2(mol)
Vo2(đktc)=n.22,4=0,2.22,4=4,48 (l)
