bài 3:
a, 2Na + 2H2O -> 2NaOH + H2
2 2 2 1
nNa = m/M = 46/23 = 2 mol => nH2 = 1 (mol); nNaOH = 2 (mol)
b, VH2 = n x 22,4 = 1 x 22,4 = 22,4 (lit)
c, mNaOH = n x M = 2 x (23+16+1) = 80 (gam)
đáp số:
2Na + 2H2O -> 2NaOH + H2
VH2 = 22,4 lit
mNaOh = 80 gam