nNa = 11.5/23 = 0.5 mol
2Na + 2H2O --> 2NaOH + H2
0.5_____________0.5_____0.25
mH2O = 189 (g)
mdd sau phản ứng = 11.5 + 189 - 0.5 = 200 g
mNaOH = 0.5*40 = 20 g
C%NaOH = 20/200*100% = 10%
D NaOH = mdd/V = 200/189=1.06 g/ml
nNa = \(\frac{11,5}{23}=0,5\) \(mol\)
2Na + 2H2O ---> 2NaOH + H2
0,5 0,5 0,25
mH2O = 189 g
mdd (sau pư) = 11,5 + 189 - 0,25.2 = 200g
mNaOH = 0,5.40 = 20 g
C% = \(\frac{20}{200}.100\%\) = 10%
DNaOH = \(\frac{200}{189}\) = 1,06 g/ml
2Na + 2H2O -> 2NaOH + H2
0,5 .....................0,5...........0,25 (mol)
nNa = 0,5(mol)
mdd = 11,5 + 189 - 0,25.2 = 200 (g)
C% = \(\frac{0,5.40}{200}.100\%=10\%\)
D = \(\frac{m}{V}=\frac{200}{189}\approx1,06\left(\frac{g}{ml}\right)\)