Question

[Trang Huynh]

giải giúp mình câu 60 với ạ

Answer 1

[๖ۣۜDũ๖ۣۜN๖ۣۜG]

\(n_{CO_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

\(n_{Na_2CO_3}=\dfrac{3,18}{106}=0,03\left(mol\right)\)

Bảo toàn C: n_C = 0,09 (mol)

Bảo toàn Na: n_Na = 0,06 (mol)

\(n_{O_2}=\dfrac{6,72.20\%}{22,4}=0,06\left(mol\right)\)

Theo ĐLBTKL: m_X + m_O2 = m_CO2 + m_H2O + m_Na2CO3

=> m_H2O = 0,54 (g)

=> \(n_{H_2O}=\dfrac{0,54}{18}=0,03\left(mol\right)\)

Bảo toàn H: n_H = 0,06 (mol)

=> \(\left\{{}\begin{matrix}\%C=\dfrac{12.0,09}{4,44}.100\%=24,33\%\\\%H=\dfrac{1.0,06}{4,44}.100\%=1,35\%\\\%Na=\dfrac{0,06.23}{4,44}.100\%=31,08\%\\\%O=100\%-24,33\%-1,35\%-31,08\%=43,24\%\end{matrix}\right.\)

=> B