a) \(n_{NaOH}=0,25.0,5=0,125\left(mol\right)\)
PTHH: 2Na + 2H2O ---> 2NaOH + H2
0,125<------------0,125
=> mNa = 0,125.23 = 2,875 (g)
b) \(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)
PTHH: 2Na + 2H2O ---> 2NaOH + H2
2------------------>2--------->1
=> VA = 1.22,4 = 22,4 (l)
mdd B = 1000 + 46 - 2 = 1044 (g)
=> \(C\%_{NaOH}=\dfrac{2.40}{1044}.100\%=7,66\%\)
\(D_{dd.NaOH}=\dfrac{1044}{966}=1,08\left(g/ml\right)\)
a) $2Na + 2H_2O \to 2NaOH + H_2$
Ta có : $n_{Na} = n_{NaOH} = 0,25.0,5 = 0,125(mol)$
$m_{Na} = 0,125.23 = 2,875(gam)$
b)
$n_{Na} = \dfrac{46}{23} = 2(mol)$
Theo PTHH : $n_{H_2} = \dfrac{1}{2}n_{Na} = 1(mol)$
$V_A = 1.22,4 = 22,4(lít)$
$n_{NaOH} = n_{Na} = 2(mol)$
$m_{dd\ B} = m_{Na} + m_{H_2O} - m_{H_2} = 46 + 1000 -1.2 = 1044(gam)$
$C\%_{NaOH} = \dfrac{2.40}{1044}.100\% = 7,66\%$
$D_{dd\ B} = \dfrac{m}{V} = \dfrac{1044}{966} = 1,081(gam/ml)$
