Đáp án:
Giải thích các bước giải:
Thí nghiệm 1 :
$n_{Na} = \dfrac{m_1}{23}(mol)$
$2Na + 2H_2O \to 2NaOH + H_2$
Theo PTHH :
$n_{NaOH} = n_{Na} = \dfrac{m_1}{23}(mol)$
$\to m_{NaOH} = \dfrac{m_1}{23}.40 = \dfrac{40m_1}{23}(gam)$
$n_{H_2} = 0,5n_{Na} = \dfrac{m_1}{46}(mol)$
Sau phản ứng :
$m_{dd} = m_{Na} + m_{H_2O} - m_{H_2}$
$= m_1 + x - \dfrac{m_1}{46}.2 = \dfrac{22m_1}{23} + x(gam)$
$\to C\%_{NaOH} = \dfrac{ \dfrac{40m_1}{23} }{ \dfrac{22m_1}{23} + x}.100\% = \dfrac{40m_1}{22m_1+23x}.100\%= y\%$(1)
Thí nghiệm 2 :
$n_{Na_2O} = \dfrac{m_2}{62}(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = \dfrac{m_2}{31}(mol)$
$\to m_{NaOH} = \dfrac{40m_2}{31}(gam)$
$m_{dd} = m_{Na_2O} + m_{H_2O} = m_2 + x(gam)$
$\to C\%_{NaOH} = \dfrac{ \dfrac{40m_2}{31}}{m_2 + x}.100\% = \dfrac{40m_2}{31m_2 + 31x}.100\%= y\%(2)$
Vì $(1) = (2)$ nên :
$ \dfrac{40m_1}{22m_1 + 23x} = \dfrac{40m_2}{31m_2 + 31x}$
$⇔ \dfrac{m_1}{22m_1 + 23x} = \dfrac{m_2}{31m_2 + 31x}$
$b)$
Với $m_1 = 4,6 ; m_2 = 6,9$ thì :
$ \dfrac{4,6}{22.4,6 + 23x} = \dfrac{6,9}{31.6,9 + 31x}$
$⇔ x = 17,74$
Suy ra :
$C\%_{NaOH} = \dfrac{40m_2}{31m_2 + 31x}.100\% = y\%$
$⇔ \dfrac{y}{100} = \dfrac{40.6,9}{31.6,9 + 31.17,74}$
$⇔ y = 36,13$
