Sửa đề: 5,475 gam HCl
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{H_2}=n_{Zn}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\cdot\dfrac{5,475}{36,5}=0,075\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,075\cdot65=4,875\left(g\right)\\V_{H_2}=0,075\cdot22,4=1,68\left(l\right)\end{matrix}\right.\)