nNa=0.23/23=0.01mol

nO2=2.24/22.4=0.1mol =>mO2=0.1*32=3.2gam

PTPU: 4Na+O2->2Na2O

 Xét: 0.01/4<0.1/1

=>Na hết, O2 dư

=> tính theo nNa

nO2=1/4nNa=0.25*0.01=0.0025mol

mO2=0.0025*32=0.08gam

mO2 dư=3.2-0.08=3.12gam

Theo PT ta có: nNa2O=1/2nNa=0.005mol

=>mNa2O=0.005*62=0.31gam

P2O5: đi phôt pho pentaoxit - thuộc ôxit axit

Đáp án:

$n_{Na}=\dfrac{0,23}{23}=0,01(mol)$

$n_{O_2}=\dfrac{2,24}{22,4}=0,1(mol)$

$4Na+O_2\to 2Na_2O$

$\dfrac{0,01}{4}<\dfrac{0,1}{1}\to$ Sau phản ứng $Na$ hết $,O_2$ dư

$\to n_{O_2}$ phản ứng $=\dfrac{0,01}{4}(mol)$

$\to n_{O_2}$ dư $=0,1-\dfrac{0,01}{4}=\dfrac{39}{400}(mol)$

$\to m_{O_2}$ dư $=32\times \dfrac{39}{400}=3,12(g)$

$n_{Na_2O}=\dfrac{0,01.2}{4}=\dfrac{1}{200}(mol)$

$\to m_{Na_2O}=\dfrac{1}{200}\times 62=0,31(g)$

$Na_2O: natri \; oxit$

$Oxit\; bazo$