a)
ms = 600g, c% = 9,8%
stim c% = mdx100/ms
=> md = msxc%/100 = 600x9,8/100 = 58,8g H2SO4
M.H2SO4 = 2x1+32+4x16 = 98 g/mol
M.NaOH = 23+16+1 = 40 g/mol
58,8g md g n moli m g
H2SO4 + 2NaOH --> Na2SO4 + 2H2O
98 2x40 1 2x18
=> md = 58,8x2x40/98 = 48 g NaOH
din c%.NaOH => ms = mdx100/c%
= 48x100/20 = 240 g sol. NaOH utilizata
b)
=> n = 58,8x1/98 = 0,6 moli Na2SO4
c)
M.H2O = 2x1+16 = 18 g/mol
m = 58,8x2x18/98 = 21,6g H2O