Question

Answer 1

\(n_{NaOH}=\dfrac{6}{40}=0,15\left(mol\right)\)

Có: m_MgO : m_FeO : m_Na2O = 2 : 2 : 1

=> \(\left\{{}\begin{matrix}\%m_{MgO}=40\%\\\%m_{FeO}=40\%\\\%m_{Na_2O}=20\%\end{matrix}\right.\)

PTHH: Na₂O + H₂O --> 2NaOH

           0,075<------------0,15

=> \(m_{Na_2O}=0,075.62=4,65\left(g\right)\)

=> \(m_A=\dfrac{4,65.100}{20}=23,25\left(g\right)\)

Answer 2

\(Na_2O+H_2O\rightarrow2NaOH\\ Tacó:n_{NaOH}=\dfrac{6}{40}=0,15\left(mol\right)\\ n_{Na_2O}=\dfrac{1}{2}n_{NaOH}=0,075\left(mol\right)\\ \Rightarrow x=m_{Na_2O}=0,075.62=4,65\left(g\right)\\ \Rightarrow2x=m_{MgO}=m_{FeO}=9,3\left(g\right)\\ \Rightarrow m_A=9,3.2+4,65=23,25\left(g\right)\)

Answer 3

nNaOH = 6/40 = 0,15 (mol)

PTHH: 2Na + 2H2O -> 2NaOH + H2

nNa = nNaOH = 0,15 (mol)

=> mNa = 0,15 . 23 = 3,45 (g)

=> mA = 3,45 . 2 + 3,45 . 2 + 3,45 = 17,25 (g)