Đáp án:
m% al= 2,7/6,3× 100%= 42,9%
m%mg= 3,6/6,3×100%= 57,1%
Giải thích các bước giải:
n$H_2$= 6,72/22,4=0,3 (mol)
ptpu
2Al+ 6HCl--> 2$AlCl_3$+ 3$H_2$
x<-------------------------------------3x/2(mol)
Mg +2HCl--> $MgCl_2$+ $H_2$
Y<-------------------------------------y (mol)
=> $\left \{ {{3x/2+y=0,3} \atop {27x+ 24y=6,3}} \right.$
<=> $\left \{ {{x=0,1} \atop {y=0,15}} \right.$
=> m al= 0,1× 27= 2,7(g)
m mg= 0,15×24=3,6(g)
m% al= 2,7/6,3× 100%= 42,9%
m%mg= 3,6/6,3×100%= 57,1%