\({n_{{H_2}}} = \dfrac{{5,04}}{{22,4}} = 0,225(mol)\)
Cách 1:
\(Mg\,\,\,\,\, + \,\,\,\,\,{H_2}S{O_4} \to MgS{O_4} + {H_2} \uparrow \)
24 gam 1 mol
x gam a mol
=> a = \(\dfrac{{x}}{{24}}\)
\(2Al\,\,\,\,\,\, + \,\,\,\,\,3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2} \uparrow \)
2 x 27 gam 3 mol
(4,5-x) gam b mol
=> \(b = \dfrac{{(4,5 - x) \times 3}}{{27 \times 2}}\)
\(\dfrac{x}{{24}} + \dfrac{{(4,5 - x) \times 3}}{{27 \times 2}} = \dfrac{{5,04}}{{22,4}} = 0,225\)
\(x = 1,8gam \to {m_{Mg}} = 1,8(gam);{m_{Al}} = 4,5 - 1,8 = 2,7(gam)\)
\(\% {m_{Al}} = 60\% ,\% {m_{Mg}} = 40\% \)
Cách 2:
\(Mg\,\,\,\, + \,\,\,\,{H_2}S{O_4} \to MgS{O_4} + {H_2} \uparrow \)
x mol x mol
\(2Al\,\,\,\,\, + \,\,\,\,\,3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2} \uparrow \)
y mol \(\dfrac{{3y}}{2}\) mol
Ta có phương trình: 24x + 27y = 4,5 (I)
\(x + \dfrac{{3y}}{2} = 0,225\) (II)
Giải (I) và (II), tìm được x và y.
