mng ơi giúp mình vs ạ đnag hấp

Đáp án + Giải thích các bước giải:

Phản ứng xảy ra:

$2Na(s)+2H_2O(l) \to 2NaOH(aq)+H_2(g)$

$n_{Na}=\dfrac{4,6}{23}=0,2(mol)$

$n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,2=0,1(mol)$

$n_{NaOH}=n_{Na}=0,2(mol)$

$m_{\text{dd NaOH}}=m_{Na}+m_{H_2O}-m_{H_2}=4,6+195,6-0,1.2=200(g)$

$C\%_{NaOH}=\dfrac{0,2.40}{200}.100\%=4\%$

$V_{NaOH}=\dfrac{200}{1,2}=\dfrac{500}{3}(ml)=\dfrac{1}{6}(l)$

$C^{M}_{NaOH}=\dfrac{0,2}{\dfrac{1}{6}}=1,2(M)$

`2Na+2H_2O->2NaOH+H_2`

`n_{Na} = {4,6}/{23} = 0,2\ (mol)`

`Theo\ pt:n_{H_2} = 1/2n_{Na} = 1/2 xx 0,2 = 0,1\ (mol)`

`m_{dd\ sau} = 4,6 + 195,6 - ( 0,1 xx 2 ) = 200\ (g)`

`Theo\ pt:n_{NaOH} = n_{Na} = 0,2\ (mol)`

`-> C%_{NaOH} = {0,2 xx 40}/{200} xx 100 = 4%`

`V_{NaOH} = {200}/{1,2} = 500/3(ml) = 1/6\ (l)`

`-> C_{M\ NaOH} = {0,2}/{1/6} = 1,2` $\rm M$