a) nNa = 0,84 mol
2Na + 2H2O ➝ 2NaOH + H2
0,84---------------------------0,42
nH2 = 0,42 mol => V = 9,408 lít
b) nNaOH = nNa = 0,84 mol
=> mO = 13,44 gam => mA ≃ 17,9 gam
a) \(n_{Na}=\frac{19,32}{23}=0,84\left(mol\right)\)
\(PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\)
________0,84____________________0,42(mol)
\(V_{H_2}=0,42.22,4=9,408\left(l\right)\)
b) \(m_{H_2O}=0,84.18=15,12\left(g\right)\)