Koliko se mililitara vodonika, racunato pod normalnim uslovima, oslobadja rastvaranjem 0,54 g aluminijuma u sumpornoj kiselini? (Al-27)Resenje : 672
Postupak ?
2Al+3H2SO4->Al2(SO4)3+3H20,54g......................................x l2*27.......................................3*22,4
x=0,672l=672ml
powered by Drupal
2Al+3H2SO4->Al2(SO4)3+3H20,54
2Al+3H2SO4->Al2(SO4)3+3H2
0,54g......................................x l
2*27.......................................3*22,4
x=0,672l=672ml