Question

calculati :
a. numarul de anioni(ioni negativi) continuti in 130 kg de hidroxid de alumiliu
b.numarul de cationi(ioni pozitivi) continuti in 8.56 kg hidroxid feric

Answer 1

a.

Al(OH)3 = hidroxid de aluminiu

M.Al(OH)3 = 27+(16+1)3 = 78 g/mol

1 mol ...................... 78 g

x moli ................... 130000 g

= 1666,67 moli Al(OH)3

1 mol Al(OH)3 ................ 3x6,022x10la23  anioni (OH)-

1666,67 moli ...................... n = 3,011x10la 27 anioni (OH)-

b.

Fe(OH)3 = hidroxid feric

M.Fe(OH)3 = 56+(16+1)3 = 107 g/mol

1 mol Fe(OH)3 ......................... 107 g

x moli ..................................... 8560 g

= 80 moli

1 mol Fe(OH)3 ............... 1x6,022x10la23 cationi Fe(3+)

80 moli .............................. n = 4,82x10la25 cationi Fe(3+)