Đáp án:
Giải thích các bước giải:
a)$n_{H_2}=\frac{11,2}{22,4}=0,5(mol)$
Gọi $\left \{ {{n_{Al}=a(mol)} \atop {n_{Fe}=b(mol)}} \right.$
$⇒27a+56b=16,6$
$⇔56b=16,6-27a$
$⇔b=\frac{16,6-27a}{56}$
$2Al+6HCl→2AlCl_3+3H_2↑$
$a:3a:a:1,5a$
$Fe+2HCl→FeCl_2+H_2↑$
$b:2b:b:b$
$∑n_{H_2}=1,5a+b=0,5$
$⇒1,5a+\frac{16,6-27a}{56}=0,5$
$⇒a=0,2$
$⇒b=\frac{16,6-27.0,2}{56}=0,2$
$⇒\left \{ {{m_{Al}=0,2.27=5,4(g)} \atop {m_{Fe}=0,2.56=11,2(g)}} \right.$
b)$V_{dd HCl}=\frac{0,2.3+0,2.2}{0,5}=2(l)$
c)$C_{ddHCl}=\frac{1.36,5}{2000.1,05}=0,017$%
Xin hay nhất!!!