Ta có:

\(\left\{{}\begin{matrix}m_{Na}=5,4.42,593=2,3\left(g\right)\\m_{Na2O}=5,4-2,4=3,1\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Na}=\frac{2,3}{23}=0,1\left(mol\right)\\n_{Na2O}=\frac{3,1}{62}=0,05\left(mol\right)\end{matrix}\right.\)

\(PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\)

__________ 0,1___________0,1________0,05

\(n_{H2}=0,05\left(mol\right)\Rightarrow V_{H2}=0,05.22,4=1,12\left(l\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,05______________0,1

\(n_{NaOH}=0,1+0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{NaOH\left(tao.thanh\right)}=0,2.40=8\left(g\right)\)

2Na+2H2O->2NaOH+H2

0,5-----0,5-----------0,5----0,25

Na2O+H2O->2NaOH

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

n H2=0,25 mol

=>m Na =0,5.23=11,5g

=>m Na2O=6,2g=>n Na2O=0,1 mol

=>m NaOH=0,7.40=28g

=>VH2O=0,6.22,4=13,44l

giúp mình với

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

0,5                                        0,25

a)\(m_{Na}=\left(\right)\cdot23=11,5g\)

\(m_{Na_2O}=17,7-11,5=6,2g\)

b)\(n_{Na_2O}=\dfrac{6,2}{62}=0,1mol\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,1                           0,2

\(\Rightarrow\Sigma n_{bazo}=0,5+0,2=0,7mol\Rightarrow m_{NaOH}=0,7\cdot40=28g\)

c)\(m_{H_2O}=\left(0,5+0,1\right)\cdot18=10,8g\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)

Anh có làm rồi nha!

a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Na}=2n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Na}=0,3.23=6,9\left(g\right)\)

\(\Rightarrow m_{Na_2O}=13,1-6,9=6,2\left(g\right)\)

b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

Theo PT: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=0,5\left(mol\right)\)

Ta có: m _{dd sau pư} = 13,1 + 200 - 0,15.2 = 212,8 (g)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,5.40}{212,8}.100\%\approx9,4\%\)

\(n_{H_2O}=\dfrac{14,4}{18}=0,8mol\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

0,8         0,8             0,8                 ( mol )

\(m_{Na}=0,8.23=18,4g\)

\(m_{NaOH}=0,8.40=32g\)

Số mol của nước là:
n_H2O=14,4/18=0,8(mol)
PTHH: Na+H2O→NaOH+1/2H₂
            0,8    0,8    0,8     0,4   ( mol)
a) Thể tích khí Hidro tạo thành(ĐKTC) là:
V_H2=0,4*22,4=  8,96(l)
b) Khối lượng Natri là:
m_Na=0,8*23 =18,4(g)
Khối lượng Bazơ tạo thành sau phản ứng là:
m_NaOH=0,8*40=32(g)