\(PTHH:Fe+2HCl->FeCl_2+H_2\)
0,15-->0,3------->0,15----->0,15 (mol)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(V_{H_2\left(dktc\right)}=n\cdot22,4=0,15\cdot22,4=3,36\left(l\right)\)
\(m_{FeCl_2}=n\cdot M=0,15\cdot\left(56+71\right)=19,05\left(g\right)\)


