PTHH: 4Na + O2 -to-> 2Na2O
Ta có : \(n_{Na}=\dfrac{6,2}{23}=\dfrac{31}{115}\left(mol\right)\)
Theo PTHH và đb, ta có:
\(n_{Na_2O}=\dfrac{\dfrac{2.31}{115}}{4}=\dfrac{31}{230}\left(mol\right)\\ n_{O_2}=\dfrac{\dfrac{31}{115}}{4}=\dfrac{31}{460}\left(mol\right)\)
a) \(m_{Na_2O}=\dfrac{31}{230}.62\approx8,357\left(g\right)\)
b) \(V_{O_2\left(đktc\right)}=\dfrac{31}{460}.22,4\approx1,51\left(l\right)\)
PTHH: \(4Na+O_2\underrightarrow{t^o}2Na_2O\)
\(n_{Na_2O}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.\)\(\dfrac{6,2}{23}=\dfrac{31}{230}\left(mol\right)\)
\(n_{O_2}=\dfrac{1}{4}.n_{Na}=\dfrac{1}{4}.\)\(\dfrac{31}{115}=\dfrac{31}{460}\left(mol\right)\)
\(m_{Na_2O}=\dfrac{31}{230}.62=\dfrac{91}{115}\left(g\right)\)
\(V_{O_2}=\dfrac{31}{460}.22,4=\dfrac{868}{575}\left(l\right)\)
