Question

problema 5, va rog! dau coroana.

Answer 1

270 g Al impur ......................... 100%

m.pur ...................................... 90%

= 243 g Al pur

243g    md

2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

2x27    3x98

=> md = 243x3x98/2x27

= 1323 g H2SO4

stim ca c% = mdx100/ms

=> ms = mdx100/c%

= 1323x100/20 = 6615 g sol. H2SO4

ro = p = ms/Vs

=> Vs = ms/p = 6615/1,104

= 5991,85 mL sol. H2SO4