Question

Va rog sa ma ajutati aceste calcule la mate​

Answer 1

1)

100% ................... 398,87 kg NaCl impura

88% ................... mp = 351 g NaCl pura

351g        md g

2NaCl + H2SO4 --> Na2SO4 + 2HCl

2x58,5      98

=> md = 98x351/2x58,5 = 294 g H2SO4

c% = mdx100/ms => ms = mdx100/c%

= 294x100/49 = 600 g sol. H2SO4

2)

ms = 800g, c% = 20%

c% = mdx100/ms => md = msxc%/100

= 800x20/100 = 160 g NaOH

                160g           a g              b g

FeSO4 + 2NaOH --> Fe(OH)2 + Na2SO4

                 2x40         90              142

=> a = 160x90/2x40 = 180 g Fe(OH)2

=> b = 160x142/2x40 = 284 g Na2SO4