Đáp án:
Giải thích các bước giải:
1.
$n_{HCl}=2n_{H_2}=2y$
$BTKL:$
$m_{hh}+m_{HCl}=m_{muối}+m_{H_2}$
$⇒x+36,5.2y=z+2y$
$⇔x+73y-2y-z=0$
$⇔x+71y-z=0$
2.
a)$n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)$
Gọi $\left \{ {{n_G=a(mol)} \atop {n_{Al}=b(mol)}} \right.$
$G+H_2SO_4→GSO_4+H_2↑$
$a:a:a:a$
$2Al+3H_2SO_4→Al_2(SO_4)_3+3H_2↑$
$b:1,5b:0,5b:1,5b$
$n_{H_2SO_4}=n_{H_2}=0,4(mol)$
$BTKL:$
$m_A+m_{HCl}=m_{\text{muối}}+m_{H_2}$
$⇒m_{\text{muối}}=7,8+0,4.36,5-0,4.2=21,6(g)$
b)$V_{H_2SO_4}=\dfrac{0,4}{2}=0,2(l)$
Xin hay nhất!!!