\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.1...................................0.1\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(m_{Fe\left(tt\right)}=0.1\cdot\dfrac{56}{90\%}=6.22\left(g\right)\)
Fe + 2HCl -> FeCl2 + H2
1 2 1 1
nH2 = 2,24/22,4 = 0,1 mol => nFe = 0,1 mol
mFe = 0,1 x 56 = 5,6 (gam)
mFe phản ứng = 5,6 x 90% = 5,04 (gam)