a) \(4Na+O_2\underrightarrow{t^0}2Na_2O\)

b)\(n_{Na}=\frac{5.52}{23}=0.24\left(mol\right)\)

\(n_{Na_2O}=\frac{2}{4}\cdot n_{Na_2O}=\frac{2}{4}\cdot0.24=0.12\left(mol\right)\)

\(m_{Na_2O}=0.12\cdot62=7.44\left(g\right)\)

c)\(n_{O_2}=\frac{1}{4}\cdot n_{Na}=\frac{1}{4}\cdot0.24=0.06\left(mol\right)\)

\(\Rightarrow V_{O_2}=0.06\cdot22.4=1.334\left(l\right)\)

\(\Rightarrow V_{kk}=V_{O_2}\cdot5=1.334\cdot5=6.72\left(l\right)\)

d)\(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)

\(n_{KMnO_4}=\frac{2}{1}\cdot n_{O_2}=\frac{2}{1}\cdot0.06=0.12\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=0.12\cdot158=18.96\left(g\right)\)

a)

\(4Na+O_2\rightarrow2Na_2O\)

b)

\(n_{Na}=\frac{3,45}{23}=0,15\left(mol\right)\)

\(\rightarrow n_{Na2O}=\frac{0,15}{2}=0,075\left(mol\right)\)

\(m_{Na2O}=0,075.62=4,65\left(g\right)\)

c)

\(n_{O2}=\frac{0,15}{4}=0,0375\left(mol\right)\)

\(V_{kk}=0,0375.22,4.5=4,2\left(l\right)\)

d)

\(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)

0,075_________________________0,0375

\(m_{KMnO_4}=0,075.158=11,85\left(g\right)\)

\(a) 4Na + O_2 \xrightarrow{t^o} 2Na_2O\\ b)n_{Na} = \dfrac{4,6}{23} = 0,2 < 2n_{Na_2O} = 2.\dfrac{12,4}{62}=0,4\)

(Vô lí)

PTHH: 4Na + O₂ -t^o-> 2Na₂O

Ta có: \(n_{Na}=\frac{0,2}{4}=0,05\left(mol\right)\)

Thể tích khí O₂ cần dùng (đktc):

\(V_{O_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)

b) Ta có: \(n_{O_2}=0,05\left(mol\right)\)

PTHH: 2KMnO₄ -t^o-> K₂MnO₄ + MnO₂ + O₂

Theo PTHH và đề bài, ta có:

\(n_{KMnO_4}=2.0,05=0,1\left(mol\right)\)

Khối lượng KMnO₄:

\(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)

a, PTHH: 4Na+O2---> 2Na2O (1)

2KMnO4----> K2MnO4+MnO2+O2 (2)

Ta có: nNa=4,6/23=0,2 mol

nO2= 1/4. nNa=1/4.0,2=0,05 mol

=> VO2=0,05.22,4=1,12 (l)

b, Ta có: nO2= 0,05 mol

Theo PTHH (2) ta có:

nKMnO4=2/1. nO2= 2/1.0,05=0,1 mol

=> mKMnO4= 0,1.158=15,8 (g)

n_Na2O = \(\dfrac{23}{310}\) ( mol )

Na₂O + H₂O \(\rightarrow\) 2NaOH

\(\Rightarrow\) m_NaOH = \(\dfrac{23}{310}\).40 = \(\dfrac{92}{31}\) (g)

\(4Na+O_2\underrightarrow{t^o}2Na_2O\)

\(\Rightarrow\) m_Na = \(\dfrac{23}{155}\).23 = \(\dfrac{529}{155}\) (g)

4Na+O2->2Na2O

0,14375-----------0,07

nNa=4,6\32=0,14375 mol

=>mNa2O=0,07.62=4,34g

Na2O+H2O->2NaOH

0,07----------------0,14 mol
=>mNaOH=0,14.40=5,6g

=>C%=2,8%

4Na+O2to-->2Na2O4

0,2---------------0,1

nNa=4,6\23=0,2(mol)

-->mNa2O=0,1.62=6,2(g)

c, PT: Na2O+H2O→2NaOH

0,1-------------0,2

=>mNaOH=0,2.40=8(g)

-->C%NaOH=8\200.100%=4%

a)

nNa = 11,5 : 23 = 0,5 mol

4Na + O₂  →  2Na₂O

Theo tỉ lệ phương trình => nO₂ phản ứng = 1/4nNa = 0,5 : 4 = 0,125 mol

=> VO₂ phản ứng = 0,125.22,4 = 2,8 lít.

b)

Na₂O  +   H₂O  →  2NaOH

nNa₂O = 1/2 nNa = 0,25 mol

=> nNaOH = 2nNa₂O = 0,5 mol

<=> C_NaOH = 0,5 : 0,25 = 2M. Và A thuộc loại hợp chất bazơ.