a) Fe + 2HCl → FeCl2 + H2
0,15 →0,3 →0,15 →0,15
nFe= \(\frac{8,4}{56}\)= 0,15 (mol)
VH2= 0,15.22,4=3,36 (l)
b) mctHCl=0,3.36,5=10,95 (g)
mddHCl=\(\frac{10,95.100}{3,65}\)= 300 (g)
c) VHCl= \(\frac{m_{dd}}{D}\)=\(\frac{300}{1,125}=267\left(ml\right)\)
d) dd A: FeCl2
mFe=0,15.56=8,4 (g)
mddFeCl2=8,4+10,95=19,35 (g)
mctFeCl2=0,15.127=19,05 (g)
C%FeCl2=\(\frac{19,05.100}{19,35}=98,44\%\)